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A TEACHER’S GUIDE TO EUCLID’S ELEMENTS

Parts of a Proof or Demonstration

In the Elements, a proposition contains these parts:

I. Enunciation (1) The words in italics just below the proposition number (2) Universal terms are used to claim that a geometric item can be constructed

(QEF) or that some geometric property is true (QED). II. Given (the “setting out”)

(1) Provides the particular—figure, angle, line, etc.—from the enunciation which the demonstration takes for granted.

III. Specification (“to do” or “to demonstrate”) (1) Stated in particular terms what is to be shown or made from the given. (2) Often introduced with “I say that…” in QED or “thus it is required…”in QEF.

IV. Construction (1) Extension of the original figure (given) to what is needed to complete the

demonstration or making of a figure. V. Proof (demonstration)

(1) Consists in steps and reasons for those steps. (a) Steps: Claims that lead from the given to the conclusion in a logically

rigorous manner. They must be true and follow one another necessarily. (b) Reasons: Provides claim(s) that show why each step is true. Reasons will

often be appeals to definitions, postulates, common notions, and/or prior propositions.

(c) Demonstrations Quia (QEF) and Demonstrations Propter Quid (QED). VI. Conclusion

(1) Restatement of the enunciation that establishes the universal as true through the particular proof.

Indirect Proof in Euclid’s Elements

Euclid makes frequent use of reductio ad absurdum (reduction to the absurd) proofs in his Elements. Instead of proving some proposition or enunciation is true by direct demonstration, reductio proves that the enunciation or that some proposition is true by assuming the opposite or the contradictory is true and showing that this leads to an absurd result. If the contradictory or opposite of the original proposition leads to an absurdity, the original enunciation or proposition must be true, which means that we have proven our claim indirectly by showing that its negation leads to an absurd consequence.

Euclid will always begin a reductio by saying something like “Suppose not…; “if possible…”; simply “if not…”

Here is an example of translating Euclid’s reductios into our two-column proof form:

From I.6

Proposition I.6 (Partial converse of I.5)

E: If in a Δ 2 <s are = to one another, the sides which subtend the = <s will also be =. G: ΔABC: <ABC = <ACB S: AB = AC (1) Per impossible: If AB does not = AC, one is greater.

Let AB > AC (1) A.R.A.A. (2) Cut from AB, DB = AC & join DC (2) I.3; Post. 1 (3) Since, DB = AC & BC is common; & <DBC = <ACB (3) I.5 (4) Thus, DC = AB b/c ΔDBC = ΔACB (4) I.4 (SAS) (5) But ΔDBC is a part of ΔACB, hence

ΔACB > ΔDBC. (5) C. N. 5 (6) Hence, ΔDBC = ΔACB & ΔDBC < ΔACB,

which is absurd. (6) 4, 5, R.A.A. (7) Thus, AB is not unequal to AC & must be = to it.

Therefore, etc.

Q.E.D.

When we translate the proposition into our two column proofs, we will follow Propositional Logic in our reasons column. “A.R.A.A.” stands for “assumption for the sake of reductio ad absurdum.” When we recognize that Euclid is going to utilize an indirect proof, we put per impossible: at the beginning of that step to indicate we are going to prove the proposition is true by assuming something impossible is true and showing that it leads to an absurd result. In the reasons column, we put A.R.A.A.. We then proceed to write our steps and their reasons in the same way we would in propositions that use direct proof. We do this until we come to the two things that cannot be true at once. In our example, it is shown that ΔDBC = ΔACB step (4) and then it is shown that ΔACB > ΔDBC step (5). These things cannot be true at once. Euclid rarely provides the step that explicitly shows the two things that cannot be true at once, but we will always provide that step. In our example, it is step (6). In the reason column, we give the numbers of the steps that yield the contradiction followed by R.A.A. And since this follows necessarily from the negation of our original proposition, it follows that our original specification must be true (step (7)).

Example of Two Column Proof

Proposition I.2

Enunciation: To place @ a given point [as an extremity] a straight line equal to a given SL. Given: A, a given point & BC, the given straight line. Specification: Make a SL @ point A = BC

(1) Join A to B forming AB (1) Post. 1 (2) Construct =lateral ΔDAB (2) I.1 (3) Extend SLs AE & BF in a straight

Line w/ DA, DB (3) Post. 2 (4) Make ○CGH w/ centre B and distance BC (4) Post. 3 (5) Make ○GKL w/ centre D & distance DG (5) Post. 3 (6) BC = BG (6) def. I.15 (7) DL = DG (7) def. I.15 (8) DA = DB (8) def. I.15 (9) Subtract DA from DL & DB from DG,

thus AL = BG. (9) C.N. 3 (10) Thus, AL = BC. (10) C.N. 1

Therefore, from point A, AL is made equal to the given BC. Q.E.F.

Abbreviations and Symbols

@ = at A.R.A.A. = assumption for reductio ad absurdum b/c = because C.N. = common notion Def. = definition POM = property of magnitudes Post. = postulate R.A.A. = reductio ad absurdum ras. = right angles sc. = scilicet (namely) seq. = sequitur (it follows that…) SL = straight line trap. = trapezium v = or w/ = with =lateral = equilateral =<ar = equiangular ┴ = perpendicular < = angle □ = parallelogram (including squares) ○ = circle Δ = triangle Γ = gnomon → = conditional (if p, then q) ↔ = biconditional (if p, then q & if q, then p)

Joyce’s Properties of Magnitudes

1.If not x = y, then x > y or x < y. 2.Not both x < y and x = y. 3.If not not x = y, then x = y. 4.If x < y and y = z, then x < z. 5.If x < y and y < z, then x < z. 6.If x = y and y < z, then x < z. 7.If x = y and y > z, then x > z. 7.If x < y, then x + z < y + z. 8.If not x > y, then x = y or x < y. 9.If not x < y and not x = y, then x > y. 10.If 2x = 2y, then x = y. 11.If x = y, then 2x = 2y.

Euclid’s Elements

Book I

DEFINITIONS:

I. Points and Lines (defs. I.1-4). (1) Point is that which has no part. (def. I.1) (2) Line is breadthless length. (def. I.2)

(a) Extremities are points (def. I.3) (b) Straight line lies evenly with points on itself (def. I.4)

II. Surface (defs. I.5-7).

(1) Surface is that which has length and breadth only. (def. I.5) (a) Extremities are lines (def. I.6) (b) Plane surface lies evenly w/ straight lines on itself. (def. I.7)

III. Angles (defs. I.8-12).

(1) Plane: two lines which meet one another and do not lie in an SL. (def. I.8). (2) Rectilineal: lines containing the angle are straight. (def. I.9).1 (3) Right angle from perpendicular lines: an SL set up on another SL making adjacent angles

equal. (def. I.10) (4) Obtuse: an angle greater than a right angle. (def. I.11). (5) Acute: an angle less than a right (def. I.12).

IV. Boundary and Figure (defs. I.13-14).

(1) Boundary is that which is the extremity of anything. (def. I.13) (2) Figure is that which is contained by any boundary or boundaries. (def. I.14).

V. Circles (defs. I.15-18).

(1) Circle: figure contained by one line and a point w/in it that makes all lines falling on it equal to one another. (def. I.15). (a) Center: Point within the circle making all lines falling on it equal is. (def. I.16). (b) Diameter: a bisecting SL through the center terminating in both directions at the

circumference. (def. I.17). (c) Semi-circle is a bisection from diameter, maintains center of original circle. (def. I.18).

VI. Rectilineal Figures (defs. I.19-22).

(1) Rectilineal figures: contained by SLs. (a) Trilateral: contained by three SLs. (b) Quadrilateral: contained by four SLs. (c) Multilateral: contained by more than four SLs. (def. I.19).

(2) Of Trilateral figures (a) Equilateral triangle: three sides equal.

1 Keep in mind that angles with curved lines are possible for Euclid.

(b) Isosceles triangle: two sides equal. (c) Scalene triangle: three sides unequal. (def. I.20 (sides))

(aa) Right-angled triangle: contains one right angle. (ab) Obtuse-angled: contains one obtuse angle. (ac) Acute-angled: All three angles are acute. (def. I.21 (angles)).

(3) Of Quadrilateral Figures (a) Square: both equilateral and right-angled. (b) Oblong: right angled but not equilateral. (c) Rhombus: equilateral but not right-angled. (d) Rhomboid: neither equilateral nor right angled but has opposite sides and angles equal. (e) Trapezia: all quadrilaterals other than these. (def. I.22).

VII. Parallel SLs

(1) Parallel SLs: straight lines in the same plane produced indefinitely in both directions will not meet in either direction. (def. I.23).

POSTULATES:

1. To draw an SL from any point to any point.

2. To produce a finite SL continuously in an SL.

3. To describe a circle w/ any center and distance.

4. All right angles are equal to one another.

5. An SL falling on two SLs making two interior angles on the same side < two right angles, if the lines on that side be extended indefinitely they will meet.

COMMON NOTIONS:

1. Things which are equal to the same thing are equal to one another.

2. If equals be added to equals the wholes are equal.

3. If equals be subtracted from equals the remainders are equal.

4. Things which coincide are equal to one another.

5. The whole is greater than the part.

Proposition I.1

E: On a given SL to construct an =lateral Δ.

G: AB, a finite SL.

S: Construct an =lateral ΔABC on AB.

(1) Describe circle BCD w/ center A & radius AB. (1) Post. 3 (2) Describe circle ACE w/ center B & radius BA. (2) Post. 3 (3) Join SLs CA and CB. (3) Post. 1 (4) Since AC = AB. (4) Def. I.15 (5) And BC = BA. (5) Def. I.15 (6) Seq. AC = BC. (6) C.N. 1

Therefore, the constructed ΔABC is =lateral. Q.E.F.

Proposition I.2

E: To place a given point [as an extremity] a SL = a given SL.

G: Point A & SL BC.

S: Construct an SL @ point A = BC.

(1) Join A to B forming AB. (1) Post. 1 (2) Construct =lateral ΔDAB. (2) I.1 (3) Produce SLs AE and BF in SL w/ DA, DB. (3) Post. 2 (4) Describe ○CGH w/ center B & radius BC. (4) Post. 3 (5) Describe ○GKL w/ center D and radius DG. (5) Post. 3 (6) BC = BG; DL = DG; DA = DB. (6) Def. I.15 (7) The remaining AL = BG. (7) C.N. 3 (8) Hence, AL, BC = BG ((6)). (8) C.N. 1

Therefore, from point A, AL is made = given BC. Q.E.F.

Proposition I.3

E: Given two unequal SLs, to cut-off from the greater one = to the lesser.

G: SLs AB > C.

S: Cut off from AB the greater a SL = C the lesser.

(1) Form AD = C. (1) I.2 (2) Describe ○DEF w/ center A & radius AD. (2) Post. 3 (3) AE = AD. (3) Def. I.15 (4) C = AD. (4) I.2 (5) Hence, C = AE. (5) C.N. 1

Therefore, AE has been cut-off from AB the greater = to C the lesser. Q.E.F.

Proposition I.4 (SAS)

E: If 2 Δs have 2 sides = to 2 sides respectively & have an < contained by those 2 sides =, the 2 Δs will be congruent (equal in every respect).

G: ΔABC & ΔDEF: AB = DE, AC = DF, & <BAC = <DFE.

S: ΔABC # ΔDEF

(1) Apply ΔABC to ΔDEF coinciding @ point A on point D, then point B will coincide w/ E b/c AB = DE.

(2) Side AC will coincide w/ side DF b/c <BAC = <EDF. (3) Point C will coincide w/ point F b/c AC = DF. (4) But point B coincides w/ E, hence, base BC will coincide w/ base EF, thus, BC = EF. (4) C.N. 4 (5) <ABC coincides w/ <DEF & <ACB coincides w/ <DFE, hence these <s are =

to one another respectively. (5) C.N. 4 (6) ΔABC will coincide w/ ΔDEF, thus ΔABC = ΔDEF in every respect. (6) C.N. 4

Therefore, ΔABC # ΔDEF. Q.E.D.2

2 Contemporary geometry maintains that SAS is a postulate for reasons associated with coordinate geometry. This proof is controversial because there are no rules given to “apply” triangles to one another. But there is also no conception of a coordinate plane, hence, rules for rigid motions cannot be held against Euclid. However, there is still a question of whether some kind of rule for moving shapes about would be necessary for this proof.

Proposition I.5 (Pons Asinorum) [Divide into three sections]

E: In isosceles Δs, the base <s are =, and if the = SLs be extended the <s under the base will be =.

G: Isosceles ΔABC: AB = AC & extend SLs from BD from AB and CE from AC.

S: <ABC = <ACB & <CBD = <BCE

Section 1:

(1) Take point F @ random on BD, then cut off from AE, AG = AF (1) I.3 (2) Join FC & GB (2) Post. 1 (3) Since AF = AG, AB = AC (FA = GA & AC = AB), &

<GAF common, ΔAFC # ΔAGB; Therefore, FC = GB, <ACF = <ABG, & <AFC = <AGB. (3) I.4 (SAS)

Section 2: (<s under the base are equal)

(1) Since AF = AG & AB = AC, BF = CG (b/c AF – AB = AG – AC & AF – AB = BF, AG – AC = CG) (1) C.N. 3

(2) Thus BF = CG & FC = GB; <BFC = <CGB & share base BC, (2) I.4 (SAS) Hence, ΔBFC # ΔCGB

(3) Therefore, <FBC = <GCB (the <s under the base are =)

Section 3: (base <s are equal)

(1) Since <ABG = <ACF & <BCF = <CBG, it follows that <ABC = <ACB (b/c <ABG - <GBC = <ACF - <FCB & <ABG - < GBC = <ABC, <ACF - <FCB = < ACB) (1) C.N. 3 Therefore, in isosceles triangles the base <s are =, and if the = lines be extended the <s under the base will also be =. Q.E.D.3

3 It is important to ask whether we may prove the base angles of an isosceles triangle are equal without extending the equal straight lines.

Proposition I.6 (Partial converse of I.5)4

E: If in a Δ 2 <s are = to one another, the sides which subtend the = <s will also be =. G: ΔABC: <ABC = <ACB S: AB = AC (1) Per impossible: If AB does not = AC, one is greater. (2) Let AB > AC (2) A.R.A.A. (3) Cut from AB, DB = AC & join DC (3) I.3; Post. 1 (4) Since, DB = AC & BC is common; & <DBC = <ACB (4) I.5 (5) Thus, DC = AB b/c ΔDBC = ΔACB (5) I.4 (SAS) (6) But triangle DBC is a part of ΔACB, hence

ΔACB > ΔDBC. (6) C.N. 5 (7) Hence, ΔDBC = ΔACB & ΔDBC < ΔACB (7) 5, 6, R.A.A. (8) Thus, AB is not unequal to AC & must be = to it.

Therefore, etc.5 Q.E.D.

Proposition I.7

E: Given 2 SLs constructed on the same SL from its extremities and meeting in a point, there cannot be constructed on the same straight line on the same side, two other SLs meeting in another point and = to the former two respectively, namely the line that shares the extremity with it.6

G: Per impossible: from the extremities of AB are lines AC & AB meeting @ C; from the extremities of AB, and on the same side, are lines AD & DB meeting @ D. Let AC =AD & CB = DB. A.R.A.A.

(1) Join CD (1) Post. I.1 (2) Since AC = AD, <ACD = <ADC (2) I.5 (3) Thus, <ADC > <DCB & <CDB > <DCB (3) C.N. 5 for each (4) Since, CB = DB, <CDB = <DCB (4) I.5 (5) Hence, <CDB > <DCB & <CDB = <DCB (5) 3, 4, R.A.A.

Therefore, etc. Q.E.D.

4 First indirect proof. Cf. Indirect proof and reductio ad absurdum above. 5 N.B. This is the first time Euclid does not bother to restate the specification or the enunciation. Let students know that the et cetera refers to the specification in QEF proofs and the enunciation in QED proofs. 6 This could be made clearer by rendering it in the following way: From a finite SL, if two lines be produced from the extremities of the finite line meeting at a point, another pair of lines may not be drawn from the extremities on the same side which meet at a different point and are yet equal to the original pair respectively. There is not a specification because the proof is showing that something cannot exist. If there were a specification, it would be something like “I say that there can be no such straight lines AD and DB.”

Proposition I.8 (SSS)

E: If all 3 sides of 2 Δs are = to one another their respective <s will be = [i.e., Δs will be congruent].

G: ΔABC & ΔDEF: AB = DE, AC = DF, and base BC = EF.

S: <BAC = <EDF

(1) Apply ΔABC to ΔDEF (2) BC = EF, BA = ED, & AC = DF (2) C. N. 4 (3) If base BC coincides w/ EF, but BA does not

coincide w/ ED and AC w/ DF, but fall beside them as EG, GF, we reach the impossibility showed in I.7 (3) I.7

(4) Hence, AC must coincide w/ DF and BA w/ ED (5) Thus, <BAC coincides w/ <EDF & <BAC = <EDF (5) C.N. 4

Therefore, etc. Q.E.D.

Proposition I.9

E: To bisect a given rectilineal angle.

G: <BAC

S: To bisect <BAC

(1) Take point D @ random on AB, then cut off from AC, AE = AD. (1) I.3

(2) Join DE & on DE make EL tri. DEF, then join AF. (2) Post. 1; I.1 (3) I say AF bisects <BAC (4) Since AD = AE & AF is common & base DF = EF,

<DAF = <EAF (b/c ΔDAF # ΔEAF). (4) Def. I.20; I.8 (SSS) Therefore, <BAC has been bisected by AF. Q.E.F.

Proposition I.10

E: To bisect a given finite SL.

G: Finite SL AB

S: To bisect AB

(1) Make =lateral ΔABC on AB. (1) I.1 (2) Bisect <ACB w/ SL CD. (2) I.9 (3) I say AB has been bisected @ D. (4) Since AC = CB & CD is common & <ACD = <BCD, thus

AD = AB (b/c ΔCAD # ΔCDB). (4) Def. 20; I.4 (SAS) Therefore, AB is bisected @ point D. Q.E.F.

Proposition I.11

E: To draw a ┴ line from a given point on a given SL.

G: SL AB & point C on it.

S: To draw a line from C ┴ AB.

(1) Take D @ random on AC & make CE = CD. (1) I.3 (2) Make =lateral ΔDFE on DE. (2) I.1 (3) Join FC. I say FC ┴ AB. (3) Post. 1 (4) Since DC = CE & CF common & DF = FE, seq. ΔFDC # ΔFCE. (4) Def. I.20; I.8 (SSS) (5) Thus <DCF = <ECF & they are adjacent, thus they are right. (5) Def. I.10 Therefore CF has been drawn ┴ AB from point C on it. QEF

Proposition I.12

E: To draw a SL ┴ to a given SL from a point not on it. G: SL AB & point C not on it. S: To draw a ┴ on AB from C. (1) Take D @ random on the other side of AB & w/ centre C & distance CD describe ○EFG. (1) Post. 3 (2) Bisect EG @ H & join CG, CH, CE. (2) I.10 (3) I say CH ┴ AB. (4) Since GH = HE & HC common & CG = CE, seq. ΔCHG # ΔCHE. (4) I.8 (SSS) (5) Thus <CHG = <EHC & they are adjacent, thus they are right. (5) Def. I.10 Therefore CH has been drawn ┴ AB from point C not on it. QEF

Proposition I.13

E: If a SL falls on another SL making <s, it will make either 2 ras. or the sum of the 2<s = 2 ras [i.e., the angles will be supplementary]. G: AB falls on CD making <CBA & <ABD. S: <CBA & <ABD are right or <CBA + <ABD = 2 ras. (1) If <CBA = <ABD, they are right. (1) Def. I.10 (2) If not, draw BE from B ┴ CD. Thus <CBE & <EBD are right. (2) I.11; Def. I.10 (3) Since <CBE = <CBA + <ABE, add <EBD to each: <CBE + <EBD = <CBA + <ABE + <EBD. (3) POM: C.N. 2 (4) Again since <DBA = <DBE + <EBA, add <ABC to each: <DBA + <ABC = <DBE + <EBA + <ABC. (4) POM; C.N. 2 (5) But <CBE + <EBD = <DBE + <EBA + <ABC, hence, <CBE + <EBD = <DBA + <ABC. (5) C.N. 1 (6) But <CBE + <EBD = 2 ras., hence <DBA + <ABC = 2ras. (6) C.N. 1 Therefore etc. QED

Proposition I.14

E: If at a point on any SL 2 SLs not lying on the same side make the adjacent <s = 2 ras, the 2 SLs are collinear. G: AB @ point B on it w/ BC, BD not on the same side: <ABC + <ABD = 2 ras. S: BD is collinear w/ CB. (1) Per impossible: BD is not collinear w/ CB. Let BE be collinear w/ CB. (1) A.R.A.A. (2) Since AB stands on CBE, <ABC + <ABE = 2 ras. (2) I.13 (3) But <ABC + <ABD = 2 ras. Hence, <ABC + <ABE = <ABC + <ABD. (3) Post. 4; C.N. 1 (4) Subtract <ABC from each: <ABD = <ABE. (4) C.N. 3 (5) But <ABE is a part of <ABD, hence <ABE < <ABD. (5) C.N. 5 (6) Thus <ABE = <ABD & <ABE < <ABD, which is absurd. (6) 4, 5, R.A.A. (7) Thus BE cannot be collinear w/ CB. In the same way, we can prove that no line except BD can be collinear w/ BE.7 Therefore, etc. QED

7 For homework or classwork, make students take another line and prove that it cannot be collinear w/ BE.

Proposition I.15

E: If 2 SLs cut one another [intersect], the resulting vertical <s will be =. G: AB & CD intersect @ E S: <AEC = <DEB & <CEB = <AED. (1) Since <CEA + <AED = 2 ras. & <AED + <DEB = 2 ras, seq. <CEA + <AED = <AED + <DEB. (1) I.13; Post. 4; C.N. 1 (2) Subtract <AED from each: <CEA = <DEB. (2) C.N. 3 (3) We may do the same for <CEB & <DEA. Thus <CEB = <DEA.8 (3) I.13; Post. 4; C.N.1: C.N.3 Therefore, etc. QED Porism I.15: If 2 SLs intersect, the resulting <s @ the point of intersection = 4 ras.9

Proposition I.16

E: In any Δ, if a side be produced, the resulting exterior < is greater than either the interior opposite <s [in the Δ]. G: ΔABC w/ side BC extended to D. S: ext. <ACD > <CBA v <BAC. (1) Bisect AC @ E & join BE and extend it to F. (1) I.10; Post. 1; Post. 2 (2) Make EF = BE. (2) I.3 (3) Join FC & extend AC to G. (3) Post. 1; Post.2 (4) Since AE = EC, BE = EF, & <AEB = <FEC, seq. ΔCFE # ΔABE. (4) I.15; I.4 (5) Thus, AB =FC & <BAE = <ECF. (6) But <ECD > <ECF. (6) C.N. 5 (7) Hence, <ACD > <BAE. (7) POM (8) Similarly, if we bisect BC <ACD can be proved > <ABC. Therefore, etc. QED

8 Make students work it out. 9 Perhaps an interpolation but an ancient one.

Proposition I.17

E: In any Δ, the sum of any 2 interior <s is less than 2 ras.

G: ΔABC

S: The sum of any two interior <s of ΔABC is less than 2 ras.

(1) Extend BC to D. (1) Post. 2 (2) Since <ACD is exterior, seq. <ACD > <ABC. (2) I.16 (3) Add <ACB to each: <ACD + <ACB > <ABC + <ACB. (3) POM (4) But <ACD + <ACB = 2 ras. Hence, <ABC + <ACB < 2 ras. (4) I.13; POM (5) Similarly we can prove that <BAC + <ACB < 2 ras. & <CAB + <ABC < 2 ras. Therefore, etc. QED

Proposition I.18

E: In any Δ, the greater side is subtended by the greater angle. G: ΔABC: AC > AB S: <ABC > <BCA (1) Since AC > AB, make AD [on AC] = AB & join BD. (1) I.3; Post. 1 (2) Thus <ADB > <DCB. (2) I.16 (3) But <ADB = <ABD. Hence, <ABD > <ACB & <ABC > <ACB. (3) I.5; POM Therefore, etc. QED

Proposition I.19 (Converse of I.18)

E: In any Δ, the greater < is subtended by the greater side. G: ΔABC: <ABC > <BCA. S: AC > AB (1) Per impossible: AC = AB v AC < AB. (1) A.R.A.A. (2) Let AC = AB. Thus <ABC = <ACB & <ABC > <ACB, which is absurd. (2) I.5; 2, G., R.A.A. (3) Let AC < AB. Thus <ABC < <ACB & <ABC > <ACB, which is absurd. (3) I.18; 3, G., R.A.A. (4) Thus AC > AB. (4) POM Therefore, etc. QED

Proposition I.20

E: In any Δ, the sum of any 2 sides is greater than the remaining side. G: ΔABC S: In ΔABC, AB + AC > BC; AB + BC > AC; & BC + AC > AB. (1) Extend AB to D, make DA = AC, & join DC. (1) Post. 2; I.3; Post. 1 (2) Since DA = AC, <ADC = <ACD. (2) I.5 (3) Thus <BCD > <ADC. (3) C.N. 5 (4) Since <BCD > BDC, seq. DB > BC. (4) I.19 (5) But DA = AC, hence, AB + AC > BC (b/c AB + AC = DA + BA). (5) POM (6) Similarly we can prove AB + BC > CA & BC + CA > AB. (6) 1-6 twice mutatis mutandis Therefore, etc. QED

Proposition I.21

E: If on one side of a Δ we construct 2 SLs from its extremities that meet w/in the Δ, then the sum of the 2 SLs will be less than the 2 remaining sides of the Δ but will contain a greater angle. G: ΔABC: on BC from B & C the SLs BD, DC meeting w/in ΔABC. S: BD + DC < BA + AC but <BDC > <BAC. (1) Extend BD to E. (1) Post. 2 (2) Thus in ΔABE, AB + AE > BE. (2) I.20 (3) Add EC to each: AB + AC(i.e., AE + EC) > BE + EC. (3) POM (4) Again in ΔCED CE + ED > CD. (4) I.20 (5) Add DB to each: CE + EB(i.e., ED + DB) > CD + DB (5) POM (6) But AB + AC > EB + CE, & a fortioiri AB + AC > CD + DB. (7) Thus in the ΔCDE, <BDC > <CED. (7) I.16 (8) And in ΔABE, <CEB > <BAC. (8) I.16 (9) But <BDC > <CEB & a fortioiri <BDC > <BAC. Therefore, etc. QED

Proposition I.22

E: From 3 SLs = 3 SLs respectively to construct a Δ [thus it is necessary that the sum of any 2 of the lines is greater than the remaining line].10 G: SLs A, B, C: A + B > C; A + C > B; & B + C > A S: To construct a Δ from lines = A, B, C. (1) Make ray DE from point D. Then make DE = A; FG = B; GH = C. (1) defs. I.2-411, Post. 2; I.3 (2) W/ centre F & distance FD describe ○DKL. (2) Post. 3 (3) W/ centre G & distance GH describe ○KLH. (3) Post. 3 (4) Join KF & KG. (4) Post. 1 (5) ΔKFG is the desired Δ. (6) Since FD = FK & FD = A, seq. FK = A. (4) Def. I.15; C.N. 1 (7) Again since GH = GK & GH = C, seq. GK = C. (5) Def. I.15; C.N. 1 (8) And FG = B. Hence, FK = A; FG = B; GK = C. Therefore ΔKFG has been constructed from 3 SLs = A, B, C, respectively. QEF

Proposition I.23

E: On a given SL @ a point on it, to construct a rectilineal angle = to a give one. G: AB w/ A the point on it & <DCE. S: To construct on AB from A a rectilineal angle = to <DCE. (1) On CD, CE take points D, E respectively @ random & join DE. (1) Post. 1 (2) From 3 SLs = CD, DE, & CE respectively construct ΔAFG: CD = AF; CE = AG; DE = FG. (2) I.22 (3) Since CD = AF; CE = AG; & DE = FG, seq. <DCE = <GAF (b/c ΔCDE # ΔFGA). (3) I.8 (SSS) Therefore on AB @ A <GAF has been constructed = <DCE. QEF 10 This is not proven in the proposition but needed for the given: it follows immediately from I.20. 11 It is important for the students to see that Euclid does not want to import anything beyond his definitions, postulates, common notions, and previous proofs. While the editors do not supply the reasons by which we may construct a ray, the knowledge of definitions I.2-4 and postulate 2 suffice for its construction.

Proposition I.24

E: If 2 Δs have 2 sides = 2 sides respectively but one of the contained angles is greater than the other, they will also have the base greater than the base [namely, the one with the greater angle will have the greater base]. G: ΔABC & ΔDEF: AB = DE, AC = DF, & <A > <D. S: BC > EF (1) Since <BAC > <EDF, construct on DE @ D <EDG = < BAC. (1) I.23 (2) Make DG = AC v DF & join EG, FG. (2) I.3; Post. 1 (3) Since AB = DE, AC = DG, & <BAC = <EDF, seq. BC = EG (b/c ΔABC # ΔDGE). (3) I.4 (SAS) (4) Since DF = DG, seq. <DGF = <DFG. (4) I.5 (5) Thus <DFG > <EGF & a fortiori <EFG > <EGF (5) C.N. 5; POM (6) Since in ΔEFG <EFG > <EGF, seq. EG > EF. (6) I.19 (7) But EG = BC, hence, BC > EF. (7) POM Therefore etc. QED

Proposition I.25 (Converse of I.24)

E: Converse of I.24 is true. If 2 Δs have 2 sides = 2 sides respectively & one base greater than the other, the Δ w/ the greater base will have the greater angle contained by the = sides. G: ΔABC & ΔDEF: AB = DE, AC = DF, & base BC > EF. S: <BAC > <EDF. (1) Per impossible: <BAC = <EDF v <BAC < <EDF. (1) A.R.A.A. (2) If <BAC = <EDF, then BC = EF (b/c AB = DE & AC = DF means that ΔABC # ΔDEF). (2) I.4 (3) Hence, BC = EF & BC > EF. (3) 2, G., R.A.A. (4) If <BAC < <EDF, then BC < EF. (4) I.24 (5) Thus BC < EF & BC > EF. (5) 4, G., R.A.A. (6) Hence, <BAC > <EDF. (6) POM Therefore, etc. QED

Proposition I.26 (ASA & AAS)

E: If 2 Δs have 2 angles = and a side = to a side either adjoining the = angles or subtending one of them, the Δs will be congruent. G: ΔABC & ΔDEF: <ABC = <DEF; <BCA = <EFD; & BC = EF [for ASA]. S: AB = DE, AC = DF, & <BAC = <EDF (in other words, the Δs will be congruent). Part I. (1) Per impossible: AB > DE. (1) A.R.A.A. (2) Make BG = DE & join GC. (2) I.3; Post. 1 (3) Since BG = DE, BE = EF, & <GBC = <DEF, seq. GC = DF & <GCB = <DFE (b/c ΔGBC # ΔDEF). (3) I.4 (SAS) (4) But ex hypothesi <DFE = <BCA, hence <BCA = <GCB. (4) C.N. 1 (5) But <BCA > <GCB. (5) C.N. 5 (6) Hence, <BCA = <GCB & <BCA > <GCB. (6) 4, 5, R.A.A. (7) Thus AB = DE. (7) POM (8) But BC = EF & <ABC = <DEF, thus AC = DF & <BAC = <EDF. (b/c ΔABC # ΔDEF). (8) I.4 (SAS) Part II. (9) Let AB = DE12 (for AAS). (9) Addition to given. (10) I say AC = DF, BC = EF, & <BAC = <EDF. (10) new specification. (11) Per impossible: BC > EF. (11) A.R.A.A. (12) Make BH = EF & join AH. (12) I.3; Post. 1 (13) Since AB = DE, BH =EF, & <ABC = <DEF, seq. AH = DF & <BHA = <EFD (b/c ΔABH # ΔDEF). (13) I.4 (SAS) (14) But <EFD = <BCA, hence, <BCA = <BHA. (14) C.N. 1 (15) But <BHA > <BCA. (15) I.16 (16) Thus <BHA = <BCA & <BHA > <BCA. (16) 14, 15, R.A.A. (17) Hence, BC = EF. (17) POM (18) But AB = DE & <ABC = <DEF, hence, AC = DF & <BAC = <EDF (b/c ΔABC # ΔDEF). (18) I.4 (SAS). Therefore, etc. QED

12 Instead of “BC = EF”.

Proposition I.27

E: If a SL falling on 2 SLs make the alternate angles =, the SLs are parallel. G: EF falling on AB, CD: <AEF = <EFD. S: AB ll CD. (1) Per impossible: AB & CD if extended will meet either toward B, D v A, C. (1) A.R.A.A. (2) Extend them to B, D meeting @ G. (2) Post. 2 (3) In ΔGEF, <AEF = <EFG. But <AEF > <EFG. (3) I.16 (4) Hence, <AEF = <EFG & <AEF > <EFG. (4) G., 3, R.A.A. (5) Hence AB & CD will not meet. (6) Similarly, it can be proved that neither will they meet towards A,C. (7) Hence, AB ll CD. (5) Def. I.23 Therefore etc. QED

Proposition I.28

E: If a SL falling on 2 SLs makes the corresponding angles (ext. & int. opp.) = or interior angles on the same side are supplementary, the 2 SLs are parallel. G: EF falling on AB & CD: <EGB = <GHD & <BGH + <GHD = 2 ras. S: AB ll CD (1) Since <EGB = <GHD & <EGB = <AGH, seq. <GHD = <AGH. (1) I.15; C.N. 1 (2) But <GHD & <AGH are alternate, hence AB ll CD. (2) I.27 (3) Since <BGH + <GHD = 2 ras & <AGH + <BGH = 2 ras., seq. <AGH + <BGH = <BGH + <GHD. (3) I.13; C.N. 1 (4) Subtract <BGH from each: <AGH = <GHD. (4) C.N. 3 (5) But <AGH & <GHD are alt., hence AB ll CD. (5) I.27 Therefore etc. QED

Proposition I.29 (Converse of I.27 & I.28)

E: Converse of I.27 & I.28 is true. If the lines are parallel these angle properties are true. G: EF falls on AB ll CD. S: <AGH = <GHD (alt.); corresponding <EGB = <GHD; & <BGH, <GHD are supplementary. (1) Per impossible: <AGH > <GHD v <GHD > <AGH. Let <AGH > <GHD. (1) A.R.A.A. (2) Add <BGH to each: <AGH + <BGH > <GHD + <BGH. (2) POM (3) But <AGH + <BGH = 2 ras, thus <GHD + <BGH < 2 ras. (3) I.13; POM (4) Hence, AB, CD if extended will meet. But they are ex hypothesi parallel. (4) Post. 5 (5) Thus AB, CD will both meet and not meet if extended. (5) Def. I.23; 4, G., R.A.A. (6) Hence, <AGH = <GHD (alt.). (7) Since <AGH = <EGB, seq. <EGB = GHD (corr.). (6) I.15; C.N. 1 (8) Add <BGH to each: <EGB + <BGH = <GHD + <BGH. (7) C.N. 2 (9) But <EGB + <BGH = 2 ras., thus <GHD + <BGH = 2 ras. (supp.). (8) I.13; C.N. 1 Therefore etc. QED

Proposition I.3013

E: SLs parallel to the same SL are parallel to one another. G: AB ll EF & CD ll EF. S: AB ll CD (1) Let GK fall on all three. (2) Since GK falls on AB ll AF, seq. <AGK = <GHF (alt). (2) I.29 (3) Since GK falls on EF ll CD, seq. <GHF = <GKD (corr). (3) I.29 (4) Hence, <AGK = <GKD and they are alternate. Thus, AB ll CD. (4) C.N. 1; I.27 Therefore etc. QED

13 The ending of this proposition seems truncated. It does not provide a reason for AB ll CD which is I.27, and it does not give us a “therefore etc” or a restatement of the universal. Hence, I have provided both.

Proposition I.31

E: To construct a SL parallel to a given SL through a given point [in this case, not on the given SL]. G: Point A & BC. S: To draw a line through A parallel to BC. (1) Take point D @ random on BC & join AD. (1) Post. 1 (2) Construct <DAE on DA @ A = <ADC. (2) I.23 (3) Produce AF in a SL w/ EA. (3) Post. 2 (4) Thus <EAD = <ADC (alt.), hence, EAF ll BC. (4) I.27 Therefore, through A, EAF has been constructed ll BC. QEF

Proposition I.32

E: In any Δ, if one side be extended, the resulting exterior angle = the sum of the interior opposite angles w/in the Δ, and the sum of the interior angles of a Δ = 2 ras. G: ΔABC w/ BC extended to D. S: <ACD = <CAB + <ABC & <CAB + <ABC + <BCA = 2 ras. (1) Draw CE, through C, ll AB. (1) I.31 (2) Since AB ll CE w/ AC falling on them, seq. <BAC = <ACE (alt.). (2) I.29 (3) Since AB ll CE w/ BC falling on them, seq. <ECD = <ABC (corr.) (3) I.29 (4) But <ACE = <BAC, hence <ACD = <BAC + <ABC (b/c <ACD = <ECD + <ACE). (4) POM; C.N. 1 (5) Add <ACB to each: <ACD + <ACB = <BAC + <ABC + <ACB. (5) C.N. 2 (6) But <ACD + <ACB are supp. (6) I.13 (7) Hence, <BAC + <ABC + <ACB = 2 ras. (7) C.N. 1 Therefore etc. QED

Proposition I.33

E: The SLs joining = & parallel lines are themselves = & parallel. G: AB ll CD & AB = CD. AC, BD joining AB and CD. S: AC = BD & AC ll BD. (1) Join BC. Since AB ll CD, <ABC = <BCD (alt.). (1) Post. 1; I.29 (2) Since AB = CD & BC common & <ABC = BCD, seq. AC = BD & <ACB = <CBD (b/c ΔABC # ΔDCB). (2) I.4 (SAS) (3) Since BC makes <ACB = <CBD (alt.), AC ll BD. (3) I.27 Therefore etc. QED

Proposition I.34

E: In parallelogrammic areas the opposite sides and angles are = & the diameter bisects the areas. G: □ADCB & BC its diameter. S: The opposite sides and angles of □ABCD are = & BC bisects its area. (1) Since AB ll CD, <ABC = <BCD (alt.). (1) I.29 (2) Since AC ll BD, <ACB = <CBD (alt.). (2) I.29 (3) Thus AB = CD, AC = BD, & <BAC = <CDB (b/c 1, 2, & BC common, thus, ΔABC # ΔBDC). (3) I.26 (ASA) (4) Since <ABC = <BCD & <CBD = <ACB, seq. <ABD = <ACD. (4) C.N. 2 (5) But <BAC = <CDB, hence, the opposite sides and angles are =. (6) The diameter bisects the area. (7) Since AB = CD, BC common, & <ABC = <BCD, seq. ΔABC # ΔDCB.14 (7) I.4 Therefore, BC bisects □ADCB. QED

Proposition I.35

E: Parallelograms which are on the same base and in the same parallels15 are =.

G: □ABCD & □EBCF on the same base BC & in the same parallels AF ll BC. S: □ABCD = □EBCF. (1) Since ABCD is a parallelogram, AD = BC. (1) I.34 (2) For the same reason, EF = BC. Hence, AD = EF. (2) I.34; C.N. 1 (3) And since DE is common, AE = DF (b/c AE = AD + DE FD = EF + DE & EF = AE). (3) C.N. 2 (4) But AB = DC, thus AE = FD, AB = DC, & <FDC = <EAB (corr.). (4) I.34; I.29 (5) Thus EB = FC (b/c ΔEAB # ΔFDC). (5) I.4 (SAS) (6) Subtract ΔDGE from each Δ: trap. ABGD = trap. EGCF. (6) C.N. 3 (7) Add ΔGBC to each trap.: □ABCD = □EBCF. (7) C.N. 2 Therefore etc. QED

14 Euclid assumes that we know that if a triangle is congruent to another triangle that the area of these triangles will be equal. Though this seems obvious, students often mistake the difference between “congruent” and “equal in some respect.” This is due not only to the arguably inconsistent use of the word in contemporary geometry but also because Euclid never gives us the term “congruent” to distinguish it from “equal in some respect”. Instead, he relies on context to show which he means. In this case, he makes the inference from the fact that they have equal sides and equal angles to their equality with respect to area. Since many propositions henceforth will contain both concepts, it is important to consistently remind students of this difference. 15 One should point out that in the same parallels is equivalent to being “under the same height.” This was pointed out as early as Proclus.

Proposition I.36

E: Parallelograms which are on = bases are = to one another. G: □ABCD & □EFGH on BC = FG respectively in AH ll BG. S: □ABCD = □EFGH (1) Join BE & CH. (1) Post. 1 (2) Since BC = FG, FG = EH, hence, BC = EH. (2) I.34; C.N. 1 (3) But BC ll EH, hence, EB = HC & EB ll HC. (3) I.33 (4) Thus EBCH is a □. (4) I.34 (5) And □EBCH = □ABCD. (5) I.35 (6) But, for the same reason, □EBCH = □EFGH. Hence, □ABCD = □EFGH. (6) I.35; C.N. 1 Therefore etc. QED

Proposition I.37

E: Δs on the same base & in the same parallels are = to one another. G: ΔABC & ΔDBC on the same base BC & in AD ll BC. S: ΔABC = ΔDBC (w/ respect to area) (1) Extend AB in both directions to E, F. (1) Post. 2 (2) Through B draw BE ll CA & through C draw CF ll BD. (2) I.31; I.31 (3) Thus □EBCA = □DBCF (b/c they are on the same base and in the same parallels). (3) I.35 (4) Hence also, ΔABC = ½ □EBCA & ΔDBC = ½ □DBCF. (4) I.34; I.34 (5) [But halves of equal things are =], hence, ΔABC = ΔDBC. (5) POM Therefore etc. QED

Proposition I.38

E: Δs which are on = bases and in the same parallels are = to one another. G: ΔABC & ΔDEF on bases BC = EF & in BF ll AD. S: ΔABC = ΔDEF (with respect to area). (1) Extend AD in both directions to G, H. (1) Post. 2 (2) Through B draw BG ll CA & through F draw FH ll DE. (2) I.31; I.31 (3) Thus □GBCA = □DEFH. (3) I.36 (4) Moreover ΔABC = ½ □GBCA & ΔFED = ½ □DEFH. (4) I.34; I.34 (5) [But halves of equal things are =]. Thus ΔABC = ΔFED. (5) POM Therefore etc. QED

Proposition I.39

E: = Δs which are on the same base & on the same side are in the same parallels [under the same height]. G: ΔABC = ΔDBC on base BC on the same side. S: ΔABC, ΔDBC are in the same parallels. (1) Join AD. I say AD ll BC. (1) Post. 1 (2) Per impossible: AD is not ll BC. (2) A.R.A.A. (3) Hence, through A draw AE ll BC & join EC. (3) I.31; Post. 1 (4) Thus ΔABC = ΔEBC. (4) I.37 (5) But ΔABC = ΔDBC, hence ΔEBC = ΔDBC. (5) C.N. 1 (6) But ΔEBC < ΔDBC. (6) C.N. 5 (7) Hence, ΔEBC = ΔDBC & ΔEBC < ΔDBC, which is absurd. (7) 5, 6, R.A.A. (8) Thus AE is not parallel to BC and we can prove in the same way that no other line except AD ll BC. Therefore etc. QED

Proposition I.41

E: If a □ shares a base w/ a Δ & is in the same parallels, the □ = 2Δ. G: □ABCD & ΔEBC share base BC & are in BC ll AE. S: □ABCD = 2ΔEBC. (1) Join AC. Hence, ΔABC = ΔEBC. (1) Post. 1; I.37 (2) But □ABCD = 2ΔABC. (2) I.34; C.N. 2 (3) Hence, □ABCD = 2ΔEBC. (3) C.N. 1; POM Therefore etc. QED

Proposition I.42

E: To construct in a given rectilineal angle a □ = a given Δ. G: ΔABC & <D. S: Construct in <D a □ = ΔABC. (1) Bisect BC @ E & join AE. (1) I.10; Post. 1 (2) On EC @ E construct <CEF = <D. (2) I.23 (3) Through A draw AG ll EC & through C draw CG ll EF. (3) I.31; I.31 (4) Hence, FECG is a parallelogram. (5) Since BE = EC, seq. ΔABE = ΔAEC. (5) I.38 (6) Thus ΔABC = 2ΔAEC. (6) POM (7) But □FECG = 2ΔAEC. Thus □FECG = ΔABC. (7) I.4116; C.N. 1 (8) And <CEF = <D is in □FECG. Therefore □FECG = ΔABC [with respect to area] has been constructed in <CEF = <D. QEF

Proposition I.43

E: In any □, the complements of the □s about the diameter are =. G: □ABCD w/ diameter AC w/ □EH & □FG about AC. Leaving □BK, □KD the complements. S: □BK = □KD. (1) Since □ABCD has diameter AC, ΔABC = ΔACD. (1) I.34 (2) Again, □EH has diameter AK, hence, ΔAEK = ΔAHK. (2) I.34 (3) For the same reason, ΔKFC = ΔKGC. (3) I.34 (4) Hence, ΔAEK + ΔKGC = ΔAHK + ΔKFC.17 (4) C.N. 2 (5) But ΔABC = ΔACD, hence, the remaining □BK = □KD. (5) C.N. 3 Therefore etc. QED

16 This reason is curiously omitted in the text. 17 B/c ΔAEK = ΔAHK & ΔKGC = ΔKFC (Kentucky Fried Triangle).

Proposition I.44

E: To a given SL to apply in a given rectilineal angle a □ = a given Δ. G: AB, ΔC, & <D. S: To apply to AB in <D a □ = ΔC. (1) Make □BEFG = ΔC in <EBG = <D: BE is in a SL w/ AB. (1) I.42 (2) Draw FG through to H & draw through A AH ll BG v EF. (2) Post. 2; I.31 (3) Join HB. Since HF falls on AH ll EF, seq. <AHF + <HFE = 2 ras. (3) I.29 (4) Hence, <BHG + <GFE < 2 ras. & will meet. (4) C.N. 5; Post. 5 (5) Thus HB will meet FE. Extend them to meet @ K. (5) Post. 2 (6) Through K draw KL ll EA v FH and extend HA, GB to L,M. (6) I.31; Post. 2 (7) Then HLFK is a □ & HK its diameter. Hence, □LB = □BF. (7) I.43 (8) But □BF[□BEFG] = ΔC, hence, □LB = ΔC. (8) C.N. 1 (9) And since <GBE = <D, seq. <ABM = <D. (9) I.15; C.N. 1 Therefore □LB = ΔC in <ABM = <D has been constructed. QEF

Proposition I.45 E: To construct in a given rectilineal angle a □ = to a given rectilineal figure. G: trap. (or whatever) ABCD & <E. S: To construct a □ in <E = trap. ABCD. (1) Join DB & construct □FH = ΔABD in <HKF = <E. (1) I.42 (2) Apply □GM = ΔDBC to GH in <GHM = <E. (2) I.44 (3) Since <E = <HKF & <GHM, seq. <GHM = <HKF. (3) C.N. 1 (4) Add <KHG to each: <HKF + <KHG = <GHM + <KHG. (4) C.N. 2 (5) But <HKF + <KHG are supp. Hence, <GHM + <KHG are supp. (5) I.29; C.N.1 (6) Hence, KH is in a SL w/ HM. (6) I.14 (7) Since HG falls on KM ll FG, seq. <MHG = <HGF (alt.) (7) I.29 (8) Add <HGL to each: <MHG + <HGL = <HGF + <HGL. (8) C.N. 2 (9) But <MHG + <HGL are supp. Hence, <HGF + <HGL are supp. (9) I.29; C.N. 1 (10) Thus FG is in a SL w/ GL. (10) I.14 (11) Since FK ll HG & FK = HG; HG ll ML & HG = ML, seq. FK ll ML. (11) I.34; I34; C.N. 1; I.30 (12) And KM = FL & KM ll FL. Hence, KFLM is a □. (12) I.33 (13) Since ΔABD = □FH & ΔDBC = □GM, trap. ABCD = □KFLM. Therefore □KFLM = trap. ABCD in the <FKM = <E has been constructed. QEF

Proposition I.46 E: On a given SL to describe a square. G: AB S: To describe a square on AB. (1) Draw AC ┴ AB from A. (1) I.11 (2) Make AD = AB & through D draw DE ll AB & through B Draw BE ll AD. (2) I.3; I.31; I.31 (3) Thus ADEB is a □. Hence, AB = DE & AD = BE. (3) I.34 (4) But AB = AD, hence, BA = AD = DE = BE & □ADEB =lateral. (4) C.N. 1 (5) But it is also right angled. (6) For since AD falls on AB ll DE, seq. <BAD + <ADE are supp. (6) I.29 (7) But <BAD is right, hence <ADE is also right. (7) Post. 4; C.N. 3 (8) Hence, <ABE & <BED are right. (8) I.34 (9) Thus □ADEB is right-angled & =lateral. Hence, □ADEB is a square. (9) Def. I.22 Therefore □ADEB has been described on AB. QEF

Proposition I.47

E: In right Δs, the □ on the side subtending the right angle is = to the sum of the □s. on the sides containing the right angle.

G: ΔABC; <BAC is right.

S: BC = AB + AC .

1. Describe □BDEC on BC, □GB on AB, & □HC on AC (1) I.46 2. Through A, draw AL ll CE & join AD, FC. (2) I.31; Post. 1 3. <BAC & <BAG are ras., hence CA is in a SL w/ AG (3) I.14 4. <BAC & <CAH are right, hence BA is in a SL w/ AH (4) I.14 5. Since <DBC = <FBA; Add <ABC to each. <DBC + <ABC

= <FBA + <ABC (5) Post. 4; C.N. 2 6. <DBC + <ABC = <DBA & <FBA + <ABC = <FBC, thus

<DBA = <FBC. (6) C.N. 2 7. Since DB = BC & FB = BA; AB = FB & BD = BC; &

<ABD = <FBC; seq. ΔABD # ΔFBC. Hence, AD = FC. (7) Def. I.22; I.4 (SAS)

8. □BL = 2ABD (8) I.41 9. □GB = 2FBC (9) I.41 10. But the doubles of equals are =, hence 2ABD = 2FBC, (10) POM.; C.N. 1

□BL = □GB 11. We proceed in the same fashion by joining AE,

Hence, □CL = □HC (11) Post. 1 12. Thus □BDEC = □GB + □HC (12) C.N. 2 13. Hence, BA + AC = BC

Therefore, etc. Q.E.D.

Book III

Proposition III.1

E: To find the centre of a ○. G: ○ABC S: To find the centre of ○ABC. (1) Draw AB @ random through ○ABC & bisect AB @ D. (1) I.10 (2) Draw DC from D ┴ AB. (2) I.11 (3) Extend DC to E and bisect CE @ F. (3) Post. 2; I.10 (4) F is the centre. (5) Per impossible: F is not the centre, take G as the centre. (5) A.R.A.A. (6) Join GA, GD, & GB. (6) Post. 1 (7) Since AD = DB, DG common, & GA = GB, seq. <ADG = <GDB. (7) def. I.15; I.8 (SSS)18 (8) But when a SL set on another SL makes adjacent <s =, those <s are right. Thus <GDB is right. (8) Def. I.10 (9) But <FDB is also right. Hence, <GDB = <FDB. (9) Post. 4 (10) But <FDB > <GDB. (10) C.N. 5 (11) Thus <FDB = <GDB & <FDB > <GDB, which is absurd. (11) 9, 10, R.A.A. (12) Thus G is not the centre of ○ABC. (13) Similarly, we can prove that no other point than F is the centre. Therefore F is the centre of ○ABC. QEF Porism: If a SL in a ○ is bisected at right angles, the centre is on the cutting SL.

18 Because ΔGDA # ΔGDB.

Proposition III.2

E: If 2 random points are taken on the circumference of a ○, the SL joining them will fall w/in the ○. G: ○ABC w/ A, B randomly taken on the circumference. S: The SL joining AB will fall w/in the ○. (1) Per impossible: let the SL fall outside as AEB. (1) A.R.A.A. (2) Take centre D of ○ABC & join DA, DB, & draw DFE through. (2) III.1; Post. 1 (3) Since DA =DB, seq. <DAE = <DBE. (3) I.5 (4) & since AEB is a side extended from ΔDAE, seq. <DEB > <DAE. (4) I.16 (5) But <DAE = <DBE, hence, <DEB > <DBE. (5) POM (6) But the greater < is subtended by the greater side, thus, DB > DE. (6) I.19 (7) But DB = DF, hence, DF > DE. (7) POM (8) But DF < DE. (8) C.N. 5 (9) Hence, DF > DE & DF < DE. (9) 7, 8, R.A.A. (10) Whence, the SL joing AB will not fall outside the ○. (11) Similarly, we can prove that it will not fall on the circumference. Then the SL must fall w/in the ○. Therefore etc. QED

Proposition III.319

E: If in a ○ a SL through the centre bisects a chord or secant not through the centre, it cuts it @ right angles. And if it cuts it at right angles it also bisects it. G: ○ABC w/ CD through the centre bisecting AB (not through the centre) @ F. S: CD cuts AB @ right angles. (1) Take centre E of ○ABC & join EA, EB. (1) III.1; Post. 1 (2) Since AF = FB, FE common, & EA = EB, seq. <AFE = <BFE. (2) Def. I.15; I.8 (SSS) (3) Thus <AFE & <BFE are right. (3) Def. I.10 (4) Hence, CD through the centre cuts AB @ right angles. (5) (Converse) Let CD cut AB @ right angles. (6) I say CD bisects AB. (7) With the same construction, since EA = EB, seq. <EAF = <EBF. (7) Def. I.15; I.5 (8) But <AFE = <BFE & EF common, hence ΔEAF # ΔEBF. (8) Post. 4; I.26 (AAS) (9) Hence, AF = FB. Therefore etc. QED

19 If I am not mistaken, this is the first proposition we do that contains its own converse rather than having a separate proposition for that purpose. You may, if you like, introduce the notion of a biconditional (if P, then Q & if Q, then P (in notation P → Q & Q → P or P ↔ Q)). The pattern seems to be that Euclid will not provide a separate proposition to prove a converse, if the same construction can be used for both.

Proposition III.4

E: If in a ○ 2 SLs which are not through the centre cut one another, they do not bisect one another. G: ○ABCD: 2 SLs AC, BD which are not through the centre cut one another. S: AC does not bisect BD nor does BD bisect AC. (1) Per impossible: BD bisects AC & AC bisects BD: AE = EC & BE = ED. (1) A.R.A.A. (2) Take centre F of ○ABCD & join FE. (2) III.1; Post. 1 (3) Thus <FEA is right. (3) III.3 (4) Again <FEB is right. (4) III.3 (5) Hence, <FEA = <FEB. (5) Post. 4 (6) But <FEA < <FEB. (6) C.N. 5 (7) Thus <FEA = <FEB & <FEA < <FEB, which is absurd. (5) 5, 6, R.A.A. (8) Hence AC & BD do not bisect one another. Therefore etc. QED

Proposition III.5

E: If 2 ○s cut one another, they cannot have the same centre. G: ○ABC & ○CDG cut one another @ B, C. S: ○ABC & ○CDG cannot have the same center. (1) Per impossible: Let E be the centre of both. (1) A.R.A.A. (2) Join EC & draw EFG through @ random. (2) Post. 1 (3) Since EC = EF & EC = EG, seq. EF = EG. (3) Def. I.15; C.N. 1 (4) But EF < EG. (4) C.N. 5 (5) Hence, EF = EG & EF < EG, which is absurd. (5) 3, 4, R.A.A. (6) Thus E is not the centre of both ○s. Therefore etc. QED

Proposition III.6 E: If 2 ○s are tangent, they cannot have the same centre.

G: ○ABC tangent ○CDE @ C.

S: They cannot have the same centre.

(1) Per impossible: Let F be the centre of both. (1) A.R.A.A. (2) Join FC and draw FEB through @ random. (2) Post. 1 (3) Since FC = FB & FC = FE, seq. FB = FE. (3) Def. I.15; C.N. 1 (4) But FB > FE. (4) C.N. 5 (5) Hence, FB = FE & FB > FE, which is absurd. (5) 3, 4, R.A.A. (6) Thus F cannot be the centre of both ○s. Therefore etc. QED

Proposition III.7

E: If a point other than the centre is taken on the diameter and from that point SLs fall on the ○, the line will be greatest on which the centre lies, the remainder of the same diameter will be least, & of the rest the SLs nearer the center will be greater than the more remote. Futhermore, only 2 = SLs will fall on the point (one on each side of the least). G: ○ABCD w/ centre E & diameter AD. Take any point F [except the centre] on AD & from F make FB, ,FC, FG fall on the circle. S: FA is the greatest, FD least, & FB > FC > FG. Part I. (1) Join BE, CE, & GE. (1) Post. 1 (2) Hence, EB + EF > BF. (2) I.20 (3) But AE = BE, hence, FA > BF. (3) Def. I.15; POM (4) Since BE = CE, FE common, & <BEF > <CEF, seq. FB > FC. (4) Def. I.15; C.N. 5; I.24 (5) For the same reason, FC > FG. (5) Def. I.15; C.N. 5; I.24 (6) Again, since GF + FE > EG & EG = ED, seq. GF + FE > ED. (6) I.20; Def. I.15; POM (7) Subtract FE from each: FG > FD. (7) POM (8) Hence, FA > FB > FC > FG > FD. (8) 3, 4, 5, 7 Part II. (9) From F only 2 = SLs will fall on ○ABCD (one on each side of FD). (10) On EF @ E construct <FEH = <GEF & join FH. (10) I.23; Post. 1 (11) Since GE = EH, EF common, & <GEF = <HEF, seq. ΔEFH # ΔFEG, hence, FG = FH ( 2SLs from F). (11) I.4 (SAS) (12) No other SL = FG will fall on the circle from F. (13) Per impossible: Let FK = FG so fall. (13) A.R.A.A. (14) Since FG = FK & FH = FG, FK = FH. (14) C.N. 1 (15) But FK > FH. (15) See above. (16) Thus FK = FH & FK > FH, which is absurd. (16) 14, 15, R.A.A. (17) Hence, no other line = FG from F will fall on ○ABCD. Therefore etc. QED

Proposition III.8

E: If a point is taken outside a ○ & from the point SLs be drawn through the ○, one of which is through the centre & the others @ random, then (1) the SLs which fall on the concave arc: the one through the centre is greatest, and the rest are greater to less as they are nearer to the center, (2) the SLs which fall on the convex arc: the one between the point and the diameter is least, and the rest are less which are nearer the least, (3) only 2 = SLs will fall on the circle from the point (one on each side of the least). G: ○ABC w/ point D taken outside ○ABC; Let DA, DE, DF, & DC be SLs through the circle: DA passes through the centre. S: arc AEFC (concave): DA > DE > DF > DC & arc HLKG (convex): DG < DK < DL < DH. Part I (AEFC). (1) Take centre M of ○ABC & join ME, MF, MC, MK, ML, MH. (1) III.1; Post. 1 (2) Since AM = EM, add MD to each: DA = EM + MD. (2) Def. I.15; C.N. 2 (3) But EM + MD > ED, hence DA > ED. (3) I.2020; POM (4) Since ME = MF, MD = MD, & <EMD > <FMD, seq. ED > FD. (4) C.N. 5; I.24 (5) In the same way, we can prove that FD > CD. (6) Hence, DA > DE > DF > DC. Part II (HLKG). (7) Since MK + DK > MD & MG21 = MK, seq. DG < DK. (7) I.20; Def. I.15; POM (8) Since on MD (a side of ΔMLD) 2 SLs MK, KD meet w/in it, seq. MK + DK < ML + DL. (8) I.21 (9) But MK = ML, hence, DK < DL. (9) POM (10) In the same way, we can prove DL < DH. Hence, DG < DK < DL < DH. Part III (2 = SLs one on each side of DG). (11) On MD @ M, construct <DMB = <KMD & join DB. (11) I.23; Post. 1 (12) Since MK = MB, MD common, & <KMD = <BMD, seq. DK = DB (b/c ΔKMD # ΔBMD). (12) I.4 (SAS) (13) No other line = DK can fall on ○ABC on either side of DG. (14) Per impossible: let DN = DK so fall. (14) A.R.A.A. (15) Since DN = DK & DK = DB, seq. DN = DB. (15) C.N. 1 (16) But DN > DB. (16) See above (17) Hence, DN = DB & DN > DB, which is absurd. (17) 15, 16, R.A.A. (18) Thus no more than 2 = SLs may fall from D on either side of DG. Therefore etc. QED

20 This reason is omitted from the text. 21 Because MD is composed of MG + DG which must be less that DK + MK because MG = MK, hence the remaining DG must be less than DK.

Proposition III.9

E: If a point be taken w/in a ○ & more than 2 = SLs fall from the point on the ○, that point is the centre. G: ○ABC & point D w/in it: DA = DB = DC fall on D. S: D is the centre of ○ABC (1) Join AB, BC & bisect them @ E, F respectively. (1) Post. 1; I.10 (2) Join ED, FD and extend them to the circumference @ G, K, H, L. (2) Post. 1; Post. 2 (3) Since AE = EB, ED common, & DA = DB, seq. <AED = <BED. (3) I.8 (SSS) (4) Thus <AED & <BED are right. (4) Def. I.10 (5) Since GK bisects AE & EB @ ras,, seq. the centre is on GK. (5) Por. III.1 (5) For the same reasons, the centre must be on HL. (6) GK & HL intersect only @ D, hence, D is the centre.22 Therefore etc. QED

Proposition III.10

E: A ○ does not cut another ○ at more points than two. G: Per impossible: ○ABC cuts ○DEF @ B, G, F, H. (G) A.R.A.A. (1) Join BH, BG bisected @ K, L respectively. (1) Post. 1; I.10 (2) Draw KC ┴ BH & LM ┴ BG carried to A, E respectively. (2) I.11; Post. 2 (3) Thus the centre of the ○ is on AC. (3) Por. III.1 (4) Again, the centre of the ○ is on NO. (4) Por. III.1 (5) AC & NO meet only at P, hence, P is the centre of ○ABC. (6) Similarly, we can prove that P is also the centre of ○DEF. (7) But ○s that cut one another cannot have the same center. (5) 5, 6, III.5, R.A.A. Therefore etc. QED

22 This seems to rely on the unstated but evident fact that two straight lines, which are not in a straight line with one another, may only share a single point.

Proposition III.11

E: If 2 ○s are internally tangent, the SL joining the centres if produced will fall on the point of tangency. G: ○ADE w/ centre G internally tangent @ A to ○ABC w/ centre F. S: A SL joining G to F when produced will fall on A. (1) Per impossible: Let FGH fall [passing through D rather than A]. (1) A.R.A.A. (2) Join AF, AG. (2) Post. 1 (3) Since AG + GF > AF v FH, subtract GF from each: AG > GH. (3) POM (4) But AG = GD, hence GD > GH. (4) Def. I.15; POM (5) But GD < GH. (5) C.N. 5 (6) Thus GD > GH & GD < GH, which is absurd. (6) 4, 5, R.A.A. (7) Thus the SL joining F to G will fall on A. Therefore etc. QED

Proposition III.12

E: If 2 ○s are externally tangent, the SL joining the centres will pass through the tangent point. G: ○ABC (w/ centre F) tangent to ○ADE (w/ centre G) @ A. S: The SL joining F to G will fall on the point of contact. (1) Per impossible: Let FCDG pass through [not the point of contact]. (1) A.R.A.A. (2) Join AF, AG. (2) Post. 1 (3) Hence, AF = FC & AG = GD. (3) Def. I.15 (4) Thus AF + AG = FC + GD. (4) C.N. 2 (5) Hence, the whole FG > FA + GA. (5) POM (6) But FG < FA + GA. (6) I.20 (7) Thus FG > FA + FG & FG < FA + GA, which is absurd. (7) 5, 6, R.A.A. (8) Hence, the SL joining F, G will pass through the point of contact. Therefore etc. QED

Proposition III.13

E: A ○ is not tangent to another @ points more than one, whether internally or externally tangent. G: Per impossible: ○EBFD internally tangent to ○ABCD @ D, B. (G) A.R.A.A. (1) Take centres G, H of ○ABCD, ○EBFD. (1) III.1 (2) Thus SL joining G, H will fall on BD. Let if fall as BGHD. (2) III.11; Post. 1 (3) Hence, BG = GD & BG > HD, thus BH > HD. (3) Def. I.15; C.N. 5; POM (4) And BH = HD (b/c they are radii). (4) Def. I.15 (5) Thus BH > HD & BH = HD, which is absurd. (5) 3, 4, R.A.A. (6) Therefore, ○ are not internally tangent @ points more than one. (7) Per impossible: ○ACK externally tangent ○ABCD @ A, C. (7) A.R.A.A. (8) Join AC. (8) Post.1 (9) Since A, C are random, seq. the SL joining them will fall w/in each ○. (9) III.2 (10) But the line falls w/in ○ABCD & w/out ○ACK23 which is absurd (10) Def,. III.3; 9, 10, R.A.A. (11) Thus a circle is not externally tangent to another @ points more than one. Therefore etc. QED

Proposition III.14

E: In a ○, = chords are equidistant to the centre, and the converse is true. G: ○ABCD & chords AB = CD. S: AB, CD are equidistant from the centre. Part I (AB = CD, therefore equidistant). (1) Take centre E & from it draw EF ┴ AB & EG ┴ CD; join AE, EC. (1) III.1; I.12; Post. 1 (2) Since EF ┴ AB, seq. EF bisects AB. (2) III.3 (3) Thus AF = FB, hence, AB = 2AF. (3) POM (4) Again, CG = DG, hence CD = 2CG. (4) POM (5) Since, AE = EC, seq. AE = EC . (5) Def. I.15; POM (6) But AF + EF = AE & EG + GC = EC , hence, AF + EF = EG + GC . (6) I.47; C.N. 1 (7) But AF = GC , hence, EF = EG & thus EF = EG. (7) POM; C.N. 3; POM (8) Hence, the perpendiculars are =, thus, AB, CD are equidistant from the centre. (8) Def. III.4 Part II (Converse). (9) W/ the same construction, repeat steps 3-6. (10) Since AF = GC, AB = 2AF, & CD = 2GC, seq. AB = CD. (9) POM; C.N. 1 Therefore etc. QED

23 This entails that tangent circles have cut one another, which is prohibited by Def. III.3.

Proposition III.15

E: Of SLs in a ○, the diameter is greatest & of the rest the nearer to the center is always greater than the more remote.

G: ○ABCD w/ centre E & diameter AD. Chord BC nearer to AD than FG.

S: AD > BC > FG. (1) Draw EH ┴ BC & EK ┴ FG. (1) I.11 (2) Since BC is nearer than FG, seq. EK > EH. (2) Def. III.5 (3) Make EL = EH & through L draw LM ┴ EK carried through to N. (3) I.3; I.11; Post. 2 (4) Join ME, EN, FE, & EG. (4) Post. 1 (5) Since EH = EL, BC = MN. (5) III.14 (6) Since AE = EM & ED = EN, seq. AD = ME + EN.24 (6) Def. I.15; POM (7) But ME + EN > MN & MN = BC, thus AD > BC. (7) I.20; POM (8) Since ME + EN = FE + EG & <MEN > <FEG, seq. MN > FG. (8) Def. I.15; C.N. 5; I.24 (9) MN = BC, hence, BC > FG. (9) POM (10) Thus AD > BC > FG. Therefore etc. QED

24 Because AD = ED + AE

Proposition III.16

E: The SL drawn ┴ to the diameter of a ○ from its extremity will fall outside the ○ & into the space between the SL & the circumference another SL cannot be interposed.25 G: ○ABC w/ centre D & diameter AB. S: The SL drawn from A ┴ AB will fall outside the ○. Part I (┴ falls outside) (1) Per impossible: the SL falls w/in as CA & join DC. (1) A.R.A.A.; Post. 1 (2) Since DA = DC, seq. <DAC = <ACD. (2) Def. I.15; I.5 (3) But <DAC is right, hence, <ACD is right. (3) C.N. 1 (4) Thus <DAC + <ACD = 2 ras. (5) But <DAC + ACD < 2 ras. (5) I.17 (6) Hence, <DAC + <ACD = 2 ras. & <DAC + <ACD < 2 ras. (6) 4, 5, R.A.A. (7) Thus the SL will not fall w/in the circle. (8) Similarly, we can prove that it will not fall on the circumference. Hence, it will fall outside. Part II (No SL can be interposed). (9) Let AE fall as the ┴. (9) I.11 (10) Per impossible: let FA be interposed. (10) A.R.A.A. (11) Draw, from D, DG ┴ ‘FA. (11) I.11 (12) Since <AGD is right & <DAG < ra., seq. AD > DG. (12) I.17; I.19 (13) But AD = DH, thus DH > DG. (13) Def. I.15; POM (14) But DH < DG. (14) C.N. 5 (15) Thus DH > DG & DH < DG, which is absurd. (15) 13, 14, R.A.A. (16) Hence, no SL can be interposed. Therefore etc. QED Porism: The SL drawn ┴ to the diameter from its extremity is tangent to the circle [see Def. III.2].

25 The third part of the proof can be omitted as it involves horn angles, which are no longer considered angles.

Proposition III.17

E: From a given point [not on the circumference] to draw an SL tangent to a given ○.

G: Point A, ○BCD.

S: To draw from point A a SL tangent to circle BCD.

(1) Take center E of ○BCD; join AE. (1) III.1; Post. 1 (2) W/ center E and distance AE, describe ○AFG (2) Post. 3 (3) From D draw DF ┴ EA; join EF & AB. (3) I.11; Post. 1 (4) Re-specify: I say that AB has been drawn for from A

Tangent to ○BCD. (5) Since E is the center of both ○BCD and ○AFG, EA = EF &

ED = EB. And <E is common. (5) Def. I.15 (6) Hence, ΔDEF # ΔBEA and thus base DF = AB &

<EDF = <EBA (6) I.4 SAS (7) But <EDF is right; hence, <EBA is right & since EB is a

radius, AB touches ○BCD. (7) III.16, Por. Therefore, from A, AB has been drawn tangent to ○BCD. Q.E.F.

Propositon III.18

E: If an SL is tangent to a ○, and an SL is joined from the center to the point of contact, the SL so joined will be ┴ to the tangent.

G: DE tangent to ○ABC (w/ center F) at point C, & join FC.

S: FC ┴ DE

(1) Per impossible: let FG be drawn from F ┴ DE. (1) A.R.A.A.; I.12 (2) Since <FGC is right, <FCG is acute (2) I.17 (3) Hence, FC > FG (3) I.19 (4) But FC = FB, hence FB > FG (4) P.O.M. (5) However, FB < FG (5) C.N. 5 (6) Hence, FB > FG & FB < FG, which is absurd. (6) 4, 5, R.A.A. (7) Thus FG cannot be ┴ DE (8) We can show in the same way that no other SL except FC

┴ DE. Therefore, etc. Q.E.D.

Proposition III.19

E: If a SL is tangent to a ○ and from the point of contact a SL be drawn ┴ to the tangent, the center of the ○ will be on that SL.

G: DE tangent to ○ABC @ C, & CA ┴ DE

S: I say that the center of the circle is on AC.

(1) Per impossible: let point F be the center; join CF (1) A.R.A.A.; Post. 1 (2) Since DE is tangent to ○ABC & FC has been joined

From the center to the point of contact, FC ┴ DE (2) III.18 (3) Hence, <FCE is right, but <ACE is also right. Hence, (3) Post. 4

FCE = <ACE (4) However, <ACE < right angle, hence FCE > ACE (4) C.N. 5 (5) Thus, <FCE = <ACE & <FCE > <ACE (5) 3, 4, R.A.A. (6) Whence, F is not the center of ○ABC. (7) Similarly, we can prove that no other point except a

point on AC will be the center of ○ABC. Therefore, etc. Q.E.D.

Proposition III.20

E: In a ○ the < @ the center is 2< @ the circumference when the <s have the same circumference as the base.

G: Circle ABC, <BEC @ the center, <BAC @ the circumference and let them have the same circumference BC.

S: <BEC = 2<BAC

(1) Join AE and extend to F. (1) Post.1; Post.2 (2) Since EA = EB, the <EAB = <EBA, hence

<EAB + <EBA = 2<EAB (2) Def. I.15; I.5; P.O.M. (3) But <BEF = <EAB + <EBA, hence, <BEF = 2<EAB (3) I.32; C.N. 1 (4) For the same reason <FEC = 2<EAC (5) Hence, <BEC = 2<BAC (b/c 2<EAB = <BEF &

2<EAC = <FEC; <BEF + <FEC = <BEC & <EAB + <EAC = <BAC). (5) C. N. 2

(6) Let another SL be inflected, and let there be another <BDC; Join DE and extend to G. (6) Post.1; Post.2

(7) We can prove that <GEC = 2<EDC of which <GEB = 2<EDB.

(8) Thus <BEC = 2<BDC Therefore, etc. Q.E.D.

Proposition III.22

E: The opposite <s of quadrilaterals in circles are supplementary.

G: ○ABCD & quadrilateral ABCD in it

S: The opposite <s of the quadrilateral will be supplementary.

(1) <CAB + <ABC +<BCA = 2 right angles (1) I.32 (2) <CAB = <BDC (b/c they are in the same arc BADC) &

<ACB = <ADB (b/c they are in the same arc ADCB) (2) III.21 (3) Hence, the whole <ADC = <BAC + <ACB

(b/c <BDC + <ADB = <ADC) (3) C. N. 2 (4) Add <ABC to each: <ABC + <BAC + <ACB = <ADC + <ABC (4) C. N. 2 (5) But <CAB + <ABC + <ACB are supp., hence

<ADC + <ABC are supp. (5) C. N. 1 (6) In the same way, we can prove <BAD + <DCB are supp.

Therefore, etc. Q.E.D.

Proposition III.23

E: On the same SL there cannot be constructed 2 similar and unequal segments of circles on the same side.

(1) Per impossible: on the same SL, AB let 2 similar and unequal segments of ○s ACB & ADB be constructed on the same side. Draw ACD through and join CB, DB. (1) A.R.A.A.; Post.1

(2) Since segment ACB is similar to segment ADB, <ACB = <ADB (2) Def. III.11 (3) But <ACB > <ADB (3) I.21 (4) Hence, <ACB = <ADB & <ACB > <ADB (4) 2, 3 R.A.A.

Therefore, etc. Q.E.D.

Proposition III.24

E: Similar segments of ○s on equal SLs are equal to one another.

G: Similar segments of circles AEB, CFD on SLs AB = CD.

S: Segment AEB = CFD

(1) If we apply AEB to CFD placing point A on point C and AB on CD, point B will also coincide w/ point D (b/c AB = CD), hence, the whole segment AEB will coincide w/ CFD. (1) C.N. 4

(2) If AB = CD, but per impossible AEB does not coincide w/ CFD, it will fall within, outside, or awry as CGD. (2) A.R.A.A.

(3) Hence, a circle would cut another at more points than 2, which is absurd. (3) III.10, R.A.A. (4) Thus AEB will not fail to coincide w/ CFD.

Therefore, etc. Q.E.D.

Proposition III.25

E: Given a segment of a ○, to describe the complete circle of which it is a segment.

G: seg. ABC

S: To describe the complete circle belonging to seg. ABC.

(1) Bisect AC @ D, draw DB perp. to AC; join AB (1) I.10; I.11; Post. 1 (2) <ABD is >, =, or < BAD (3) Let it be greater (<ABD > <BAD); Construct on BA <BAE = <ABD;

Draw DB through to E; join EC. (3) I.23; Post. 2; Post. 1 (4) Since <ABE = <BAE, EB = EA (4) I.6 (5) And since AD = DC & DE is common, <ADE= <CDE base AE = CE

(b/c tri. EAD = tri. EDC). (5) Post. 4; I.4 SAS (6) But AE = BE, hence BE = CE, (7) Thus, AE = EB = EC (7) C.N. 1 (8) Thus the circle drawn with center E has two or more = Sls passing

Through it. (8) III.9 (9) It is also clear that arc ABC is < a semicircle because centre E is

outside it. (9) Def. I.18 10. Similtudine, if <ABD = <BAD, DA = DB = DC (b/c the angle drawn will coincide with <BAD) and D will be the center of the circle and ABC will be a semicircle. (10) I.6; III.9; Def. I.18 11. If <ABD is < <BAD, and we construct <EAD = <ABD, the center Will fall on DB within the segment @ E. Hence, arc ABC will be > Semicircle. (11) I.23; Def. I.18 Therefore, from a given arc the complete circle has been described. Q.E.F.

Proposition III.26

E: In # circles, = <s stand on = arcs, whether they stand @ the centres or @ the circumferences.

G: ○ABC = ○DEF: <BGC = <EHF @ the centers & <BAC = <EDF @ the circumferences.

S: arc BKC = arc ELF

(1) Join BC, EF (1) Post. 1 (2) Since BG = EH, GC = HF & <G = <H, base BC = EF (2) Def. III.1; I.4 SAS (3) Since <A = <D, arc BAC is similar to arc EDF & they are on = SLs. (3) Def. III.11 (4) Similar segments on = SLs are = to one another, hence, arc BAC = EDF (4) III.24 (5) But ○ABC # ○DEF, hence, the remaining arcs BKC, ELF are

equal. (5) C. N. 3 Therefore, etc. Q.E.D.

Proposition III.27 (Converse of III.26)

E: In # circles, <s standing on = arcs are = to one another whether they stand @ the centres or @ the circumferences.

G: ○ABC # ○DEF on arcs BC = EF. Let <s BGC & EHF stand @ centres G, H, respectively and <s BAC & EDF @ the [limits of] the arcs.

S: <BGC = <EHF & <BAC = <EDF

(1) For per impossible if <BGC does not equal to EHF, one is greater. Let <BGC > <EHF. (1) A.R.A.A.

(2) On SL BG @ point G on it, make <BGK = <EHF (2) I.23 (3) Now, = <s stand on = arc, when they are at the centres

Hence, arc BK = EF. (3) III.26 (4) But arc EF = BC, hence, BK = BC (4) C.N. 1 (5) But arc BK is a part of arc BC, hence BK < BC (5) C.N. 5 (6) Thus BK = BC & BK < BC (6) 4, 5 R.A.A. (7) Therefore, <BGC = <EHF (8) And <BAC = ½ <BGC & <EDF = ½ <EHF (8) III.20 (9) Therefore, <BAC = <EDF (9) P.O.M.

Therefore, etc. Q.E.D.

Proposition III.28

E: In #, = chords cut off = arcs, the greater = to the greater and the less = to the less. G: ○ABC # ○DEF. Chord AB = DE cutting off arcs ACB, DFE the greater arcs & arcs AGB, DHE the lesser arcs. S: arc ACB = arc DFE & arc AGB = arc DHE (1) Take centres K, L & join AK, KB, DL, & LE. (1) III.1; Post. 2 (2) Since ○s are =, the radii are =. Hence, AK = DL, KB = LE

& AB = DE (2) Def. III.1 (3) Thus ΔAKB # ΔDLE, hence, <AKB = <DLE (3) I.8 (4) But = <s stand on = arcs when @ the centres, hence,

arc AGB = arc DHE. (4) III.26 (5) But ○ABC # ○DEF, hence arc ACB = arc DFE (5) C. N. 3

Therefore, etc. Q.E.D.

Proposition III.29 (Converse of III.28)

E: In # circles, = arcs are subtended by = chords.

G: circle ABC = circle DEF; cut off = arcs BGC, EHF and join BC & EF.

S: BC = EF

(1) Take centres K, L & join BK, KC, EL, & LF (1) III.1; Post. 2 (2) Since arc BGC = arc EHF, <BKC = <ELF (2) III.27 (3) BK = EL & KC = LF (& <BKC = <ELF), hence ΔKBC

is congruent to ΔLEF. Thus, BC = EF. (3) Def. III.1; I.4 Therefore, etc. Q.E.D.

Proposition III.31 E: In a circle the < in the semicircle is right; that in a greater segment acute and in the lesser segment obtuse. Furthermore, the < of the greater segment is obtuse and the angle of the lesser segment is acute. G: ○ABCD w/ diameter BC & centre E. Join BA, AC, AD, & DC. S: <BAC = right angle; <ABC < right angle (is acute); & <ADC > right angle (is obtuse).

(1) Join AE & extend BA to F. (1) Post.1; Post. 2 (2) Since BE = EA, <ABE = <BAE & since CE = EA, <ACE = <CAE (2) I.5 (3) Hence, the whole <BAC = <ABC + <ACB (3) C.N. 2 (4) But <FAC (exterior) = <ABC + <ACB (4) I.32 (5) Hence, <FAC = <BAC and therefore both are right angles. (5) C.N. 1; Def. I.10 (6) Thus <BAC in the semicircle is right. (7) Next, since in ΔABC <ABC + <BAC < 2 right angles & <BAC is right,

<ABC < right angle. (7) I.17 (8) And since <ABC is in the segment ABC, which is greater than a

Semicircle, we have shown that <ABC in segment ABC is less than a Right angle.

(9) Next, since ABCD is a quadrilateral in a circle (the opposite <s = 2 right angles) and ABC < right angle, seq. <ADC > right angle. (9) III.22 (10) And <ADC is in segment ADC < semicircle. (11) Furthermore, I say the < of the greater segment (arc ABC and AC) is greater than a right angle and the < of the lesser segment (arc ADC & AC) is less than a right angle. (12) For since < contained by BA, AC is right, the angle of the greater segment (12) C.N. 5; C.N. 5 is greater than a right angle and the < contained by AC, AF is right the lesser segment is less than a right angle. Therefore, etc. QED

Proposition III.32

E: If from a tangent line chords be drawn from the point of tangency cutting the circle, the <s made with the tangent will be = the <s in the alternate segments of the circle. G: EF is tangent to the ○ABCD @ B. Draw chord BD. S: The <s which BD makes with tangent EF will be = in alternate segments of the circle (i.e., <FBD = <BAD constructed in the segment BAD & <EBD = <DCB constructed in the segment DCB.

(1) Draw BA from B perpendicular to EF and take point C @ random on the arc BD; join AD, DC, CB. (1) I.11; Post. 1

(2) Since EF is tangent to ○ABCD @ B & BA has been drawn Perpendicular to EF, the centre of the circle is on BA. (2) III.19

(3) Hence, BA is the diameter & <ADB is right. (3) III.31 (4) Thus the remaining <BAD + <ABD = right angle (4) I.32 (5) But < ABF is also right, hence <ABF = <BAD + <ABD. (5) Post. 4; C.N. 1 (6) Subtract <ABD from each: <ABF - <ABD = <BAD - <ABD,

hence, <DBF = <BAD in the alternate segment. (6) C.N. 3 (7) Next, ABCD is a quadrilateral in a circle so opposite <s = two right <s (7) III.22 (8) But <DBF + <DBE = 2 RAs, hence <DBF + <DBE = <BAD + <BCD. (8) I.13; C.N.1 (9) But <BAD = <DBF, hence the remaining <DBE = <DCB in the

Alternate segment. (9) C.N.3 Therefore, etc. Q.E.D.

Proposition V.1

E: If any number of magnitudes (lines, planes or whatever) which are equimultiples (equal multiples) of any magnitudes = in multitude (number of magnitudes), then whatever multiple one of the magnitudes is of one, it will be a multiple of all. (n(a + b + c…) = na + nb + nc…). G: Let AB be equimultiple of E and CD be an equimultiple of F; and E, F be equal in multitude (that is E goes into AB the same number of times as F goes into CD). S: Whatever multiple of AB is of E that multiple will AB, CD also be of E, F (i.e., If AB = nE & CD = nF, then AB + CD = n(E + F).

(1) Since AB is the same multiple of E that CD is of F, as many magnitudes of E as there are in AB, so many also are there in CD of magnitude F. (1) Def. V.11

(2) Divide AB into magnitudes AG = GB = E and divide CD into CH = HD = F (2) I.3

(3) Hence, the multitude of the magnitudes AG, GB will = the multitude (3) POM of the magnitudes CH, HD (sc., 2).

(4) Since AG = E & CH = F, seq. AG + CH = E + F (4) C.N. 2 (5) In the same way, GB + HD = E + F (5) C.N. 2 (6) (AG + CH) + (GB + HD) = 2(E +F) (6) P.O.M. (7) But AG + GB = AB & CH + HD = CD, hence, AB + CD = 2(E +F) (7) C.N. 2 (8) Hence, whatever multiple of AB is of E (2, 3, 4, etc.), that multiple

Will be of AB + CD = n(E +F) (where n = 2, 3, 4, etc.). Therefore, etc. QED

Proposition V.2

E: If a 1st magnitude be the same multiple of the 2nd that a 3rd is of a 4th, & a 5th also be the same multiple of the 2nd that a 6th is of the 4th , the 1st + 5th will also be the same multiple of the 2nd that 3rd + 6th is of the 4th (ma + na = a(m + n) or more generally ma + na + pa…= a(m + n + p…)). G: AB (1st) is the same multiple of C (2nd) as the magnitude DE (3rd) is of F (4th) & BG (5th) is the same multiple of C (2nd) and EH (6th) the same multiple of F (4th). S: AB + BG = mC + nC = (m + n)C & DE + EH = mF + nF = (m + n)F

(1) AB = m(C) & DE = m(F). (AB = 3(C) & DE = 3(F)). Hence, AB = mC & DE = mF (sc., 3)

(2) BG = n(C) & EH = n(F). (BG = 2(C) & EH = 2(F)). Hence, BG = nC & EH = nF (sc., 2).

(3) Thus AG = AB + BG = mC + nC = (m+n)C & DH = DE + EH = mF + nF = (m + n)F.

(4) Hence, whatever multiple AG is of C, that multiple is DH of F. This means that for Every segment on AG there will be the same number of segments on DH. Therefore, etc. QED

Proposition V.4

E: If a first magnitude have to a second the same ratio26 as a third to the fourth (i.e., equal ratios), any equal multiples whatever of the 1st and 3rd will also have the same ratio of any equal multiples whatever of the 2nd and 4th respectively taken in corresponding order. G: A:B = C:D & E = mA, F = mC; G = pB & H = pD S: E:G = F:H (mA:pB = mC:pD) (1) K = nE & L = nF; M = qG & N = qH (2) Since E = mA & F = mC & K = nE & L=nF, K is the same equal multiple

of A as L is of C. (2) V.3 (3) For the same reason, M is the same multiple of B that N is of D. (3) V.3 (4) Since A:B = C:D and equal multiples K & D have been taken &

of B, D, other chance equal multiples have been taken M, N (i.e., qG & qH), seq. if K exceeds M, then L will exceed N; if K = M, L = N; and if K < L, L< N. (K >=< M (nE >=< qG) (4) Def. V.5

(5) But K, L are equal multiples of E, F, and M, N other chance equal multiples of G, H, hence E:G = F:H. (Since L >=< N (nF >=< qH), seq. if nE >=< qG, then nf >=< qH or K:M = L:N, but nE = mA, qG = pB, nF = pD, hence, E >=< G & F >=< H is true and E:G = F:H. (i.e, mA:pB = mC:pD). (5) Def. V.5 Therefore, etc. QED

Proposition V.7

E: Equal magnitudes have to the same ratio, as also has the same to equal magnitudes. [if a = b & c does not = a v b → a:c::b:c & c:a::c:b). G: A = B with C being unequal to A, B. S: if a = b, then a:c = b:c & c:a = c:b

(1) D = nA & E = nB, F = mC (2) Since A = B, seq. D = E. (2) C.N. 1 (3) Hence, if D >=< F then E >=< F (D:F = E:F) (4) But D,E are equal multiples of A,B, while F is another chance

multiple of C, hence A:C = B:C (nA >=< mC implies nB >=<mC). (5) I say next that C also has the same ratio to each of the magnitudes

A, B (i.e., C:A = C:B). (5) Def. V.5 (6) With the same construction we prove that D = E, so F >=<D and

F is a multiple of C, while D, E are chance equal multiples of A, B, Hence, C:A = C:B (6) Def. V.5 Therefore, etc. QED

26 First, we must remember definition V.4: A and B are in a ratio iff there is some magnitude ‘p’ and ‘q’ such that pA > B & A < qB (which means, algebraically, that we cannot compare infinitesimal numbers to regular numbers or regular numbers to infinite ones). Second, we need to understand definition V.5: For any random magnitudes ‘p’ and ‘q’ it must be the case that if pA >=< qB, then pC >=< qD.

Proposition V.9 (Converse of V.7)

E: Magnitudes that have the same ratio to the same are equal to one another, and magnitudes to which the same has the same ratio are equal (if a:c:b:c or c:a:c:b, then a = b).27 G: A:C:B:C & C:A::C:B S: A = B (1) Per impossible: If A is unequal to B, then they would not

have had the same ratio to C. (b/c only one of the three cases A:C >=< B:C can be true at once). (1) A.R.A.A; V.8

(2) But A, B do have the same ratio to C, hence, A = B. (3) Again, If A were unequal to B, C would not have the

same ratio to A, B.(b/c only one of the three cases C:A >=< C:B can be true at once). (3) V.8

(4) But C does have the same ratio to A, B, hence A = B. Therefore, etc. QED.

27 Remembering def. V.4: pA > B & A < qB (only does not work if one number is infinity or one number is infinitely small). We also need to countenance Def.V.7 (what it means for a ratio to be greater than another): if nA > mB & nC <= mD, then A:B > C:D (i.e., there are two magnitudes ‘n’ and ‘m’ that satisfy these conditions). For example, 180:85 and 125:70 (when n = 2 and m = 4 satisfies our conditions, hence 180:85 > 125:70).

Proposition V.11

E: Ratios which are the same with the same ratio are also the same with one another. (if a:b = c:d & c:d = e:f, then a:b = e:f). G: A:B = C:D & C:D = E:F S: A:B = E:F

(1) G = nA, H = nC, K = nE & L = rB, M = rD, N = rF (2) G >=< L & H >=< M (2) Def. V.5 (3) H >=< M & K >=< N (3) Def. V.5 (4) Hence, G >=< L & K >=< N (5) Since G, L are equal multiples of A, E and L, N are

Equal multiples of B, F, seq. A >=< B & E >=< F. (i.e., A:B = E:F).

Proposition V.12

E: If any number of magnitudes be proportional, as one of the antecedents is to one of the consequents, so will all (the sum of) the antecedents be to all (the sum of) the consequents. (if x1:y1 = x2:y2…= xN:yN, then (x1 + x2 + x3…xN):(y1 + y2 +y3…+yN). G: A:B::C:D::E:F S: A:B::(A + C + E):(B + D + F)

Proposition V.15

E: Parts have the same ratio as the same multiples of them taken in corresponding order (i.e., if c and f are the same parts ‘n’ of corresponding magnitudes, then c:f::nc:nf). ‘ G: AB = nC & DE = nF S: C:F::AB:DE

(1) AB = nC and DE = nF. Divide AB into AG = GH = HB = C & DE into DK = KL = LE = F. (1) I.3

(2) Hence, AG:DK::GH:KL::HB:LE (the antecedents are greater in the same ratio). (2) V.7

(3) Thus AG:DK::(AG + GH + HB):(DK + KL + LE) (3) V.12 (4) But AG + GH + HB = AB and DK + KL + LE = DE, hence,

AG:DK::AB:DE. (5) But AG = C and DK = F, hence, C:F::AB:DE (i.e., C:F::nC:nF). (5) C.N.1

Therefore etc. QED

Proposition V.16

E: If 4 magnitudes be proportional, they will also be proportional alternately (i.e., if a:b::c:d, then a:c::b:d). G: A:B::C:D S: A:C::B:D

(1) E = nA, F = nB; G = mC, H = mD (2) Since parts have the same ratio to as the same multiples of them,

A:B::E:F (2) V.15 (3) Since A:B::C:D, seq. C:D::E:F (3) V.11 (4) Same as (2) w/ G,H, hence C:D::G:H (4) V.15 (5) But C:D::E:F, hence E:F::G:H (5) V.11 (6) But V.14, hence, E >=< G implies F >=< H (E:G::F:H) (6) V.14 (7) Now E, F are equal multiples of A,B and G,H of C,D;

Thus, A:C::B:D (b/c nA:mC::nB:mD). (7) Def.V.5 Therefore, etc. QED

Book VI

DEFINITIONS:

(1) Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional.

(2) A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less.

(3) The height of any figure is the perpendicular drawn from the vertex to the base.28

28 Remember that being “under the same height” is virtually the same thing as being within the same parallels.

Proposition VI.1

E: Triangles and parallelograms which are under the same height are to one another as their bases. G: ΔABC & ΔACD; □EC & □CF under the same height. S: I say that, as the base BC:CD::ΔABC:ΔACD::□EC:□EF (1) Produce BD in both directions to H, L and BG = GH = BC(base) &

DK = KL = CD(base); join AG, AH, AK, AL. (1) Post. 2; I.3; Post. 1 (2) Since CB = BG = GH, seq. ΔABC = ΔAGB = ΔAHG (2) I.38 (3) Hence, whatever multiple HC is of BC, that multiple also is of

ΔAHC of ΔABC (i.e., HC = nBC implies ΔAHC = nΔABC). (3) (4) For the same reason, LC = mCD implies ΔALC = mΔACD. (4) (5) Hence, if HC >=< CL, then ΔAHC >=< ΔACL (HC:CL::ΔAHC:ΔACL) (5) I.3829 (6) Thus BC:CD::ΔABC:ΔACD (6) Def. V.5 (7) Since pg EC = 2ΔABC & pg FC = 2ΔACD, hence

ΔABC:ΔACD::pg EC:pg FC. (7) I.41;V.15 (8) Since BC:CD::ΔABC:ΔACD & ΔABC:ΔACD::pg EC:pg FC, seq.

BC:CD::□EC:□FC. (8) V.11 Therefore, etc. QED

29 This proposition assumes that if the triangles are on unequal bases and within the same parallels, the greater base will be the greater triangle and the lesser base lesser.

Proposition VI.2

E: If a SL be drawn parallel to one of the sides of a triangle [w/in it], it will cut the sides of a triangle proportionally; and if the sides of the triangle be cut proportionally, the line joining the points of section will be parallel to the remaining side of the triangle. G: DE parallel to BC (one of the sides of ΔABC) S: BD:DA::CE:EA (& DE will be parallel to BC). (1) Join BE & CE; Hence, ΔBDE = ΔCDE. (1) Post.1; I.37 (2) But equals have the same ratio to the same (third thing, sc. ΔADE),

Hence, ΔBDE:ΔADE::ΔCDE:ΔADE. (2) V.7 (3) But ΔBDE:ΔADE::BD:DA (b/c they are under the same height). (3) VI.1 (4) For the same reason, ΔCDE:ΔADE:CE:EA. (4) VI.1 (5) Thus, BD:DA::CE:EA (5) V.11 (6) (Given) BD:DA::CE:EA. (7) (S) DE is parallel to BC. (8) With the same construction, BD:DA::ΔBDE:ΔADE &

CE:EA::ΔCDE:ΔADE, hence, ΔBDE:ΔADE::ΔCDE:ΔADE. (8) VI.1;V.11 (9) Thus, each of the Δs BDE, CDE has the same ratio to ΔADE,

Hence, ΔBDE = ΔCDE & are on the same base DE. (9) V.9 (10) = trias. on the same base are in the same parallels. (10) I.39 Therefore, etc. QED

Proposition VI.3

E: If an < of a Δ is bisected & the SL cutting the < also cuts the base, the segments of the base will have the same ratio as the remaining sides of the Δ; & the converse is true. G: ΔABC & SL AD bisecting <BAC. S: BD:CD::BA:AC

(1) Draw CE parallel to DA & carry BA through to meet CE @ E. (1) I.31; Post. 2 (2) Since AC falls on parallels AD, EC, seq. alts. <ACE = <CAD. (2) I.29 (3) But <CAD = <BAD, hence <BAD = <ACE (3) C.N.1 (4) Again, SL BAE falls on AD, EC, hence, <BAD = <AEC (int./opp). (4) I.29 (5) But <ACE = <BAD, thus <ACE = <AEC & AE = AC (5) C.N.1; I.6 (6) Since AD parallel to CE (side of ΔBCE), seq. BD:DC::BA:AE (6) VI.2 (7) But AE = AC, hence, BD:DC::BA:AC (8) (G) BA:AC::BD:DC (9) (S) If segments of the base have same ratio to the remaining sides,

AD that cuts the base will bisect <BAC of ΔABC. (10) W/ the same construction, since BD:DC::BA:AC, also BD:DC::BA:EA (b/c AD parallel CE). (10) VI.2 (11) Further, BA:AC::BA:EA (11) V.11 (12) Thus, AC = AE & <AEC = <ACE (12) V.9; I.5 (13) But <AEC = ext. <BAD & alts. <ACE = <CAD (13) I.29 (14) Hence, <BAD = <CAD and, thus, <BAC is bisected by AD. (14) C.N. 1 Therefore, etc. QED.

Proposition VI.4 (AAA ~)

E: In = <ar Δs, the sides about the = <s are proportional, & those sides which subtend the = <s are corresponding. G: ΔABC & ΔDCE: <ABC = <DCE; <BAC = <CDE; & <ACB = <CED S: In Δs ABC, DCE the sides about the = <s are proportional & the sides which subtend the = <s correspond. (1) Place BC in SL w/ CE. Since <ABC + <ACB > 2ras & <ACB = <DEC, Hence, <ABC + <DEC > 2ras, which means BA will meet ED. (1) I.17; Post. 5 (2) Produce them to meet @ F. Since, <DCE = <ABC, BF ll CD. (2) Post. 2; I.28 (3) Again, <ACB = <DEC, hence AC ll FE. (3) I.28 (4) Hence, FACD is a □ & FA = FC & AC = FD. (4) I.34 (5) Since AC ll FE (a side of ΔFBE), seq. BA:AF::BC:CE (5) VI.2 (6) But AF = CD, thus BA:CD::BC:CE & alty. BA:BC::DC:CE (6) V.16 (7) Again, CD ll BF, BC:CE::FD:DE (7) VI.2 (8) But FD = AC, thus BC:CE::AC:DE & alty. BC:CA::CE:DE. (8) V.16 (9) Since AB:BC::DC:CE & BC:CA::CE:ED, thus, ex aequali, BA:AC::CD:DE. (9) V.22 Therefore, etc. QED

Proposition VI.5 (Converse of VI.4 (SSS ~)

E: If 2 Δs have their sides proportional, the Δs will be =<ar and will have those <s = which the corresponding sides subtend. G: Δs ABC, DEF w/ proportional sides: AB:BC::DE:EF, BC:CA::EF:FD, BA:AC::ED:DF S: ΔABC is =<ar w/ ΔDEF & <ABC = <DEF; <BCA = <EFD; <BAC = <EDF (1) On EF & @ E, F on it construct <FEG = <ABC & <EFG = <ACB (1) I.23 (2) Thus the remaining <BAC = <EGF (b/c the remaining < must make the sum of the interior <s = 2ras). (2) I.32; C.N. 3 (3) Hence, ΔABC is =<ar ΔGEF and we have the conditions for VI.4, Thus AB:BC::GE:EF. (3) VI.4 (4) But ex hypothesi AB:BC::DE:EF, hence, DE:EF::GE:EF. (4) V.11 (5) Thus DE = GE. (5) V.9 (6) For the same reason, DF = GF (7) Since DE = EG, DF = GF & EF is common, seq. <DEF = <GEF Thus ΔDEF # ΔGEF. (7) I.8 SSS (8) Hence, <DFE = <GFE & <EDF = <EGF. (9) Since <FED = <GEF and <GEF = <ABC, seq. <ABC = <DEF (9) C.N. 1 (10) For the same reason <ACB = <DFE; <BAC = <EDF and ΔABC Is =<ar to ΔDEF. (10) C.N. 1 Therefore, etc. QED

Proposition VI.6 (SAS ~)

E: If 2 Δs have one < and the sides about the = <s proportional, the Δs will be =<ar and will have theose <s = which the corresponding sides subtend. G: ΔABC, ΔDEF w/ <BAC = <EDF and the sides about the = <s proportional: BA:AC::ED:DF. S: ΔABC is =<ar to ΔDEF & <ABC = <DEF, <ACB = <DFE. (1) On DF @ D, F on it, construct <FDG = <BAC (or <EDF) & <DFG = <ACB. (1) I.23 (2) Hence, the remaining <ABC = <DGF & ΔABC =<ar ΔDGF. (2) I.32 (C.N. 3) (3) Thus BA:AC::GD:DF. (3) VI.4 (4) But ex hypothesi BA:AC::ED:DF, hence GD:DF::ED:DF. (4) V.11 (5) Seq. GD = ED. (5) V.9 (6) And DF is common, hence, ED = GD, DF = DF, and <EDF = <GDF. Thus EF = GF and ΔDEF # ΔDGF. (6) I.4 SAS (7) Thus, <DFG = <DFE & <DGF = <DEF. But <DFG = <ACB, hence, <DFE = <ACB. (7) C.N. 1 (8) And ex hypothesi <BAC = <EDF, thus the remaining <ABC= <DEF. (8) I.32 (C.N. 3) (9) Thus, ΔABC =<ar ΔDEF. Therefore, etc. QED

Proposition VI.8

E: If in a right Δ a ┴ is drawn from the ra to the base, the Δs adjoining the ┴ are ~ both to the whole and one another. G: ΔABC (right) <BAC = ra & AD ┴ BC. S: ΔABD ~ ΔADC ~ ΔABC (1) Since < BAC = <ADB and <ABC is common, seq. that the remaining <ACB = <BAD, hence ΔABC is =<ar ΔABD. (1) I.32 (C.N.3) (2) Thus BC:BA::AB:BD::AC:AD. (2) VI.4 (3) So ΔABC =<ar ΔABD and has the sides prop. Thus, ΔABC ~ ΔABD. (3) Def. VI.1 (4) In the same way we can prove, ΔABC ~ ΔADC. (5) Next, we may prove that ΔABD ~ ΔADC. (6) Since <BDA = <ADC & <BAD = <ACB, seq. the remaining <ABC = <DAC. Thus ΔABD is =<ar ΔADC. (4) VI.4 (7) Hence, BD:DA::AD:DC::BA:AC, thus ΔABD ~ ΔADC. (5) Def. VI.1 Therefore, etc. QED. Porism: If in a right Δ a ┴ is drawn from the ra to the base, the line is a mean proportional between the segments of the base. (In our proposition, this means that BC:AD::AD:DC).

Proposition VI.9

E: From a given SL to cut off a prescribed part. G: AB S: Cut off from AB a prescribed part. (1) Let the third part be that prescribed. Draw AC from A w/ AB containing <CAB. Take D at random on AC and divide AC into DE = EC = AD. (1) Post. 1; I.3 (2) Join BC and through D draw DF ll BC. (2) Post. 1; I.31 (3) Since FD ll BC (side of ΔABC), seq. CD:DA::BF:FA. (3) VI.2 (4) But CD = 2DA, thus BF = 2FA, hence BA = 3AF. (4) Def. V.5 Therefore, from AB the prescribed 3rd part has been cut off. QEF

Proposition VI.11

E: To 2 given SLs to find a 3rd proportional. G: BA, AC placed so that they contain an <. S: Thus it is required to find a third line proportional to BA, AC. (1) Extend BA, AC to D, E respectively & make BD = AC. (1) Post. 2; I.3 (2) Join BC. Draw through D, DE para. BC. (2) I.31 (3) Since BC ll DE (side of ΔADE), seq. AB:BD::AC:CE. (3) VI.2 (4) But BD = AC, thus AB:AC::AC:CE. (4) V.7 Therefore, to BA, AC we have found a 3rd proportional CE. QEF.

Proposition VI.12

E: To three given SLs to find a fourth proportional. G: SLs A, B, C S: Thus it is required to find a fourth proportional to A, B, C. (1) Make an < @ random w/ lines DE, DF containing <EDF. (2) Make DG = A, GE = B, DH= C. (2) I.3 (3) Join GH & draw EF ll GH. (3) Post. 1; I.31 (4) Since GH para. EF (a side of ΔDEF), seq. DG:GE::DH:HF. (4) VI.2 (5) But DG = A, GE = B, DH = C, hence, A:B::C:HF. Therefore, to the 3 SLs, A, B, C, a fourth proportional HF has been found. QEF

Proposition VI.13

E: To 2 given SLs to find a mean proportional. G: SLs AB, BC. S: Thus it is required to find a mean proportional to AB, BC. (1) Place AB in a SL w/ AC & describe semicircle ADC. (2) Draw BD ┴ AC & join AD, DC. (2) I.10; Post.1 (3) Since <ADC is in a semicircle it is a RA. (3) III.31 (4) Since in right ΔADC, DB is drawn from RA to the base, seq. that DB is a mean proportional to segments AB, BC. (4) Por. VI.8 Therefore, given 2 SLs AB, BC a mean proportional DB has been found. (AB:DB::DB:BC) QEF

Proposition VI.14

E: In equal and =<ar PGs the sides about the equal angles are reciprocally proportional, and the partial converse is true. G: □AB = & =<ar to □BC having <s @ B =. Make DB in a SL w/ BE and thus FB & BG will be in a SL w/ one another (I.14).30 S: The sides of □AB will be reciprocally prop. to the sides of □BC (i.e., DB:BE::BG:BF). (1) Complete □FE. (1) I.31 (2) Since □AB = □BC & □FE is some third PG, seq. AB:FE::BC:FE. (2) V.7 (3) But □AB:□FE::DB:BE & □BC:□FE::BG:BF. (3) VI.1 (4) Thus also DB:BE::BG:BF(BE is a side of BC & BF of AB) (4) V.11 (5) Thus in PGs AB, BC the sides are reciprocally prop. (6) The converse is true. (7) (G) DB:BE::BG:BF (8) But as DB:BE::□AB:□FE & GB:BF::□BC:□FE. (8) VI.1 (9) Thus also □AB:□FE::□BC:□FE. (9) V.11 (10) Hence, □AB = □BC. (10) V.9 Therefore, etc. QED

Proposition VI.15

E: In = Δs which have one < =, the sides about the equal <s are reciprocally proportional & the converse is true. G: ΔABC = ΔADE w/ <BAC = <DAE. S: CA:AD::EA:AB (1) Place CA in a SL w/ AD and thus EA will be in a SL w/ AB. Join BD. (1) I.14; Post. 1 (2) Since ΔABC = ΔADE & ΔBAD is some third, seq. CAB:BAD::EAD:BAD. (2) V.7 (3) But CAD:BAD::CA:AD & EAD:BAD::EA:AB. (3) VI.1 (4) Thus also CA:AD::EA:AB. (4) V.11 (5) (G) EA:AB:CA:AD. (6) (S) ΔABC = ΔADE. (7) Since CA:AD::EA:AB, while CA:AD::ABC:BAD & EA:AB::EAD:BAD. (7) VI.1 (8) Thus, ABC:BAD::EAD:BAD. (8) V.11 (9) Whence, ABC = ADE. (9) V.9 Therefore, etc. QED

30 Cf. Heath’s commentary, p. 218 for a short and interesting proof of this.

Proposition VI.16

E: If 4 SLs be proportional, the rectangle contained by the extremes is = the rectangle contained by the means; & the converse is true.31 G: AB:CD::E:F S: The rectangle contained by AB(F) = rectangle contained by CD(E). (1) Draw AG, CH from A, C @ RAs to SLs AB, CD & make AG = F & CH = E. (1) I.11; I.3 (2) Complete □s BG, DH. (2) I.31: I.34 (3) Since AB:CD::E:F, while E = CH & F =AG, seq. AB:CD::CH:AG. (3) POM (4) Hence, in □s BG, DH the sides about the = angles are reciprocally prop. (5) But then □BG = □DH (5) VI.14 (6) Further, BG is the rectangle AB, F & DH is the rectangle CD, E. Thus the Rectangle contained by AB, F = rectangle contained by CD, E. (7) (G) Rect. AB, F = Rect. CD, E. (8) (S) AB:CD::E:F. (9) W/ the same construction—Since rect. AB, F = rect. CD, E; rect, AB, F is BG & rect. CD, E is DH; seq. □BG = & =<ar □DH. (9) I.29; I.34; Post. 4 (10) But in = & =<ar □s the sides are reciprocally prop. Thus AB:CD::CH:AG, Hence, AB:CD::E:F. (10) VI.14; POM Therefore, etc. QED

31 From Heath p. 222, He restates the enunciation thus: “Rectangles which have their bases reciprocally proportional to their heights are equal in area; and equal rectangles have their bases reciprocally proportional to their heights.”

Proposition VI.18

E: On a given SL to describe a rectilineal figure similar and similarly situated to a given rectilineal figure. G: SL AB and rectilineal figure CE. S: To describe on AB a figure similar and similarly situated to CE. (1) Join DF. And on AB @ A, B on it, make <GAB = <FCD & <ABG = CDF. (1) Post. 1; I.23 (2) Thus the remaining <CFD = <AGB, hence ΔFCD =<ar ΔGAB. (2) I.32 (C.N.3) (3) Hence, FD:GB::FC:GA::CD:AB. (3) VI.4 (4) Again, on SL BG, @ B, G in it, make <BGH = <DFE & <GBH = <FDE. (4) I.23 (5) Thus the remaining <FED = <GHB, hence, ΔFDE =<ar ΔGBH. (5) I.32 (C.N.3) (6) Thus FD:GB::FE:GH::ED:HB (6) VI.4 (7) But it was proved that FD:GB::FC:GA::CD:AB, hence, FC:AG::CD:AB::FE:GH::ED:HB. (7) V.11 (8) Since <CFD = <AGB & <DFE = <BGH, seq. <CFE = <ABH. (8) C.N. 2 (9) Similitudine, <CDE = <ABH. (9) C.N. 2 (10) Further, since <FCD = <GAB & <FED = <GHB, seq. AH =<ar CE & the sides about their equal <s are proportional which makes AH similar to CE. (10) Def. VI.1 Therefore on AB, trap. AH has been constructed similar to trap. CE. QEF

Proposition VI.19

E: ~Δs are to one another in the duplicate ratio32 of the corresponding sides. G: ΔABC ~ ΔDEF w/ <ABC = <DEF & AB:BC::DE:EF such that BC corr. EF. (Def. V.11) S: ΔABC:ΔDEF is a duplicate ratio of BC:EF (1) Take a 3rd proportional BG to BC, EF: BC:EF::EF:BG. Join AG. (1) VI.11; Post. 1 (2) Since AB:BC::DE:EF, Thus alty AB:DE::BC:EF. (2) V.16 (3) But BC:EF::EF:BG, thus AB:DE::EF:BG. (3) V.11 (4) Thus the sides of ΔABG & ΔDEF are reciprocally prop., but Δs which Sides (about the = <s) are reciprocally prop. are =. Whence, ΔABG = ΔDEF. (4) VI.15 (5) Since BC:EF::EF:BG, BC:BG is a duplicate ratio of BC:EF. (5) Def. V.9 (6) But CB:BG::ΔABC:ΔABG. (6) VI.1 (7) Thus ΔABC:ΔABG is a duplicate ratio of BC:EF. (8) But ΔABG = ΔDEF, hence, ΔABC:ΔDEF has a ratio duplicate to BC:EF. Therefore etc. QED Porism: If 3 SLs are prop., then as the 1st magnitude is to the 3rd magnitude so the figure described on the 1st:2nd similarly described. If BC:EF::EF:BG, then BC:BG::ΔABC (fig. on BC):ΔDEF (fig. on EF).

32 A duplicate ratio is the result of compounding two equal ratios. Def. V.9 if a:b::c:d, then a:c::a :b . Here a:c is the duplicate ratio of a:b. The duplicate ratio of two numbers means the ratio of their squares. Thus the duplicate ratio of the numbers 4 and 7 is 16:49. The duplicate ratio of two straight lines is the ratio of the areas of the squares described on them. This ratio is of importance in the theory of similar plane figures, for it is proved in geometry that the ratio of the areas of two such figures is equal to the duplicate ratio of any corresponding linear dimensions. If one square has its diagonal twice as great as that of another square, its area will be four times that of the second.

Proposition VI.22

E: If 4 SLs are prop., the rectilineal figures similar and similarly33 described upon them will also be proportional; and the converse is true. G: AB:CD::EF:GH; On AB, CD describe ΔKAB ~ ΔLCD & on EF, GH describe □MF ~ □NH. S: ΔKAB:ΔLCD::□MF:□NH. (1) Take a 3rd proportional O to AB, CD & P to EF, GH (AB:CD::CD:O & EF:GH::GH:P). (1) VI.11 (2) Since AB:CD::EF:GH & CD:O::GH:P, seq. ex aequali34 AB:O::EF:P. (2) Def. V.17;V.22 (3) But AB:O::KAB:LCD & EF:P::MF:NH (3) Por. VI.19 (4) Hence, KAB:LCD::MF:NH (EF:P is the MT). (4) V.11 (5) (G) MF:NH::KAB:LCD. (6) (S) AB:CD:EF:GH (7) Per impossible: EF is not to GH::AB:CD. (7) A.R.A.A. (8) EF:QR::AB:CD (take 4th prop. to EF, AB, CD). (8) VI.12 (9) On QR construct □SR ~ □MF v □NH. (9) VI.18 (10) Since AB:CD::EF:QR & on AB, CD ΔKAB ~ ΔLCD; on EF, QR □MF ~ □SR, seq. KAB:LCD::MF:SR. (11) But by hypothesis KAB:LCD::MF:NH, thus MF:SR::MF:NH. (11) V.11 (12) Thus MF has the same ratio to SR, NH, hence SR = NH. (12) V.9 (13) But NH ~ SR, thus GH = QR.35 (14) And since AB:CD::EF:QR, while QR = GH, thus AB:CD::EF:GH. Therefore, etc. QED.

33 Euclid here seems to mean some kind of preservation of orientation. 34 If a:b::d:e & b:c::e:f, then a:c::d:f (middle terms ‘b’,’e’). 35 This step does not seem to be justified. We need proof that if two similar figures are also equal, then any pair of corresponding sides are equal (Heath p. 242). However, we may ask that if two figures are equiangular and equal in area with their sides proportional, how can the sides fail to be equal?

Proposition VI.24

E: In any □ the □s about the diameter are similar to both the whole and one another. G: □ABCD w/ diameter AC. □EG & □HK about AC. S: □EG ~ □HK ~ □ABCD (1) Since EF para. BC (side of ΔABC), seq. BE:EA::CF:FA. (1) VI.2 (2) Since FG para. CD (side of ΔACD), seq. CF:FA::DG:GA. (2) VI.2 (3) Thus BE:EA::DG:GA and componendo BA:AE::DA:GA ((BE + AE):AE::(DG + GA):GA) (3) V.11; Def. V.1436 (4) And alty. BA:AD::EA:GA. (4) V.16 (5) Thus in □s ABCD, EG the sides that contain <BAD are prop. (6) And since GF para. DC, <AFG (ext.) = <DCA (int. opp.). (5) I.29 (7) And <DAC is common to ΔADC & ΔAGF, seq. ΔADC =<ar ΔAGF. (6) I.32, C.N. 3 (8) For the same reasons, ΔACB =<ar AFE. (7) I.29; I.32, C.N.3 (9) Thus □ABCD =<ar □EG, thus AD:DC::AG:GF, DC:CA::GF:FA, AC:CB::AF:FE, CB:BA::FE:EA (9) VI.4 (AAA ~). (10) Since DC:CA::GF:FA & AC:CB::AF:FE; thus ex aequali, DC:CB::GF:FE. (10) V.22 (11) Hence, □ABCD ~ □EG. (11) Def. VI.1 (12) We may repeat the proof for □HK, and thus □ABCD ~ □HK. (13) But if □ABCD ~ □EG & □ABCD ~ □HK, then □EG ~ □HK. (12) VI.21 Therefore etc. QED

36 Componendo: If a:b::c:d, then (a + b):b::(c + d):d. Separando: If a:b::c:d, then (a – b):b::(c – d):d. The reason given in the text is V.18: If magnitudes are proportional separando they will be proportional componendo. However, it seems that we only need to know Def. V.14 to justify our componendo.

Proposition VI.25

E: To construct one & the same fig. ~ to a given rectilineal fig. & = to another given rectilineal figure. G: ΔABC (to which the construction must be similar) and trap. D (to which the figure must be equal in area). S: Thus it is required to construct a figure both ~ ΔABC and = in area to trap. D (1) Apply to BC □BE = ΔABC. (1) I.44 (2) & to □CE apply □CM = trap. D w/ <FCE = <CBL. (2) I.45 (3) Thus BC is in a SL w/ CF & LE w/ EM (b/c <BCE + <CEL = 2ras And <CEL = <FCE, hence, <BCE + <FCL = 2ras). (3) I.29; I.34; C.N. 1; I.14 (4) Take mean proportional to GH to BC, CF (BC:GH::GH:CF). (4) VI.13 (5) & on GH describe ΔKGH ~ ΔABC. (5) VI.18 (6) Since, BC:GH:GH:CF, thus BC:CF::ΔABC:ΔKGH. (6) Por. VI.19 (7) But BC:CF::□BE:□EF, thus ΔABC:ΔKGH::□BE:□EF. (7) VI.1; V.11 (8) Thus alty. ΔABC:□BE::ΔKGH:□EF.37 (8) VI.16 (9) But ΔABC = □BE, thus ΔKGH = □EF (9) V.14 (cf. (7)). (10) But □EF= trap. D, thus ΔKGH = trap. D. (10) C.N. 1 (11) Thus ΔKGH ~ ΔABC & ΔKGH = trap. D. Therefore etc. QEF

Proposition VI.26

E: If a □ be taken from another ~ □ with which it shares an <, it is about the same diameter as the whole. G: □ABCD and from it take away □AF ~ □ABCD & shares <DAB w/ it. S: □ABCD is about the same diameter w/ AF. (1) Per impossible: □ABCD is not about dia. AF but dia. AHC. (1) A.R.A.A. (2) Extend GF to H & draw HK para. AD v BC. (2) Post. 2; I.31 (3) Since □ABCD is about same diameter w/ □KG, DA:AB::GA:AK. (3) VI.24 (4) But □ABCD ~ □EG, thus DA:AB::GA:AE. (4) Def. VI.1 (5) Thus GA:AK::GA:AE. (5) V.11 (6) Thus AK = AE. But AK >AE. (6) V.9; C.N. 5 (7) Hence, AK = AE & AK > AE, which is foolishness itself. (7) 6, R.A.A. (8) Thus □ABCD must be about the same dia. w/ AF. Therefore etc. QED

37 The purpose of this step is unknown because V.14 provides us with the necessary justification to conclude the equality from ΔABC:ΔKGH::□BE:□EF.

Proposition VI.29

E: To a given SL to apply a □ = to a given rect. fig. and exceeding by a parallelogramic fig. ~ to a given one.38 G: SL AB, C given rect. fig. to which the fig. applied to AB must be = & □D to which the excess is must be ~. S: Thus it is required to apply to the SL AB a □ = to C and exceeding by a □ ~ □D. (1) Bisect AB @ E. On EB describe □BF ~ □D. (1) I.10; VI.18 (2) Construct □GH ~ □D & □GH = □BF + C. (2) VI.25 (3) Let KH corr. FL & KG corr. FE. (KH:FL::KG:FE) (3) VI.21 (4) Since □GH > □FB, seq. KH > FL & KG >FE. (4) Def.V.5 (5) Extend FL & FE: FLM = KH & FEN = KG. (5) Post. 2; I.3 (6) Complete □MN. Thus □MN = ~ □GH. (6) I.31 (7) But □GH ~ □EL (□BF), thus □MN ~ □EL. (7) VI.21 (8) Hence, □EL is about the same diameter as □MN. (8) VI.26 (9) Draw diameter FO and let the figure be described □PQ. (10) Since GH = EL + C, while GH = MN, thus MN = EL + C. (10) C.N. 1 (11) Subtract □EL from each □MN – □EL = □EL + C – □EL. Hence, gnomon39 XWV = C. (11) C.N. 3 (12) Since AE = EB, □AN = □NB = □LP (Complement). (12) I.36; I.43 (13) Add EO to each: □AN + □EO = □NB + □EO = □LP + □EO, Thus AO = gnomon VWX (LP + EO). (13) C.N. 2 (14) But gnomon VWX = C, hence, AO = C. (14) C.N. 1 (15) Since PQ ~ EL, PQ ~ D. (15) VI.24 Therefore, to AB, AO = C has been applied and exceeds C by □QP ~ D (AO = C + QP) QEF

38 Solving: ax + b/c x = S. The algebraic solution is x = - c/b ∙ a/2 + √(c/b)c/b(c/b ∙ a/4 + S). N.B.: This proposition refers to a gnomon which should be covered prior to its assignment. 39 Def. II.2: In any parallelogramic area let any parallelogram about its diameter with the two complements is a gnomon.

Proposition VI.31

E: In right angled Δs the figure on the side subtending the ra is = to the (sum of) similar and similarly described figures on the sides containing the ra. G: ΔABC: <BAC is right. S: The □BC = □BA + □AC (when □BA ~ □AC). (1) Draw from <BAC, AD ┴ BC. (1) I.12 (2) Thus ΔABD ~ ΔADC ~ ΔABC. (2) VI.8 (3) Since ΔABC ~ ΔABD, seq. CB:BA::AB:BD. (3) Def. VI.1 (4) Thus CB:BD::□CB:□BA. (4) Por. VI.19 (5) Thus also since ΔABC ~ ΔADC, seq. CB:AC::CD:CA & BC:CD::□BC:□CA. (5) Def. VI.1; Por. VI.19 (6) In addition (BD + DC):BC::(□BA + □AC):□BC (6) V.2440 (7) But BC = BD + DC, thus □BC = □BA + □AC. Therefore, etc QED

Proposition VI.33

E: In = circles <s have the same ratio as the arcs on which they stand, whether they stand @ the center or at the circumferences. G: Cir. ABC = Cir. DEF w/ <BGC @ centre G & <EHF @ centre H; And <BAC on the circumference of ABC and <EDF on circumference of DEF. S: arcBC:arcEF::<BGC:<EHF::<BAC:<EDF (1) Take consecutive arcs arcCK = arcKL = arcBC and consecutive arcs arcFM = arcMN = arcEF. (1)(I.23)(III.26)? (2) Join GK, GL, HM, & HN. Since BC = CK = KL, <BGC = <CGK = <KGL. (2) III.27 (3) pBC & p<BGC; qEF & q<EHF (3) Def. V.2 (4) Hence, If arcBL >=< arcEN, then <BGL >=< <EHN. (4) III.27 (5) Thus arcBC::arcEF::<BGC:<EHF. (5) Def.V.5 (6) But <BGC:<EHF::<BAC:<EDF (b/c <BGC = 2<BAC & <EHF = 2EDF). (6) III.20; P.O.M. (7) Thus arcBC:arc:EF::<BGC:<EHF::<BAC:<EDF. (7) V.11 Therefore etc. QED

40 If a:b::c:d & e:b::f:d, then (a + e):b::(c + f):d.

Proposition II.141

E: Given 2 SLs w/ one of them cut into any number segments, the rectangle contained by the two SLs is = the sum of the rectangles by the uncut SL and each of the segments. (Demonstrates distributive law).42 G: SLs A, BC: BC is cut @ random @ D, E. S: The rectangle contained by A(BC) = A(BD) + A(DE) + A(EC) [A(BC) = A(D+E+F))] (1) Draw from B, BF ┴ BC. Make BG = A. (1) I.11; I.3 (2) Through G draw GH parallel to BC & through D, E, C, draw DK, EL, CH parallel to BG. (2) I.31 (3) Thus □BH = □BK + □DL + □EH (4) Now □BH is the rectangle A(BC) b/c it is contained by GB, BC & BG = A (□BH = GB(BC), hence □BH = A(BC)). (4) Def. II.1 (5) □BK = A(BD) b/c it is contained by GB, BD & A = BG [□BK = GB(BD), hence □BK = A(BD)]. (5) Def. II.1 (6) □DL = A(DE) b/c it is contained by DK, DE & DK = BG = A. [□DL = DK(DE), hence □DL = A(DE)]. (6) Def. II.1; I.34 (7) □EH = A(EC) b/c it is contained by EL, EC & EL = DK = BG = A. (7) Def. II.1; I.34 (8) Thus A(BC) = A(BD) + A(DE) + A(EC). (8) C.N. 2 Therefore, etc. QED

41 Heath says of geometric algebra: “The product of two numbers was thus represented geometrically by the rectangle contained by the straight lines representing the two numbers respectively. It only needed the discovery of incommensurable or irrational straight lines in order to represent geometrically by a rectangle the product of any two quantities whatever, rational or irrational; and it was possible to advance from a geometrical arithmetic to a geometrical algebra, which indeed by Euclid's time (and probably long before) had reached such a stage of development that it could solve the same problems as our algebra so far as they do not involve the manipulation of expressions of a degree higher than the second.” (p. 372). 42 Algebraic equivalent: a(b + c + d…) = ab + ac + ad…

Proposition II.243

E: If a SL is cut @ random, the rectangles (the products) contained by the whole and both of the segments are = to the square on the whole. G: AB cut randomly @ C. S: AB(BC) + AC(AB) = AB (AB(AB)). (1) Describe □ADEB on AB & draw CF through C para. AD, BE. (1) I.46; I.31 (2) Thus □AE = □AF + □CE. (3) Now □AE = AB². □AF = AC(AD), hence, □AF = AC(AB) b/c AD = AB. (4) And □CE = BE(BC), hence □CE = AB(BC) b/c BE = AB. (5) Thus AB(AC) + AB(BC) = AB . (5) C.N. 2 Therefore, etc. QED

Proposition II.344

E: If we cut a SL @ random, the product of the whole multiplied by one of the segments is = the product of the segments and the square on that segment. G: AB cut randomly @ C. S: AB(CB) = AC(CB) + BC (1) Describe □CDEB on CB & draw ED through to F & through A draw AF para. CD, BE. (1) I.46; Post. 2; I.31 (2) Thus □AE = □AD + □CE. (3) Now □AE = AB(BE), hence □AE = AB(BC) b/c BE = BC. (3) Def. II.1 (4) And □AD = AC(DC), hence □AD = AC(CB) b/c DC = CB. & DB = CB . (5) Thus AB(CB) = AC(CB) + BC (5) C.N. 2 Therefore, etc. QED

43 Algebraic equivalent: (a + b)a + (a + b)b = (a + b) [where a = BC and b = AC & a + b = AB]. 44 Algebraic equivalent: (a + b)a = ab + a [where a = BC and b = AC & a + b = AB]

Proposition II.445

E: If we cut a SL @ random, the square on the whole is = the sum of squares on the segments plus 2x the product of the segments. G: AB cut @ random @ C. S: AB = AC + CB + 2AC(BC) (1) Describe □ADEB on AB & join BD. (1) I.46; Post. 1 (2) Through C draw CF para. AD or EB & through G draw HK para. AB or DE. (2) I.31 (3) Since CF para. AD & BD is the transversal <CGB = <ADB (3) I.29 (4) But <ADB = <ABD. Thus <CGB = <GBC, hence BC = CG. (4) I.5; I.6 (5) But CB = GK & CG = KB, hence GK = KB. Thus CGKB is equilateral. (5) I.34; C.N. 1 (6) But it is also right-angled. Since CG ll BK, <KBC + <GCB = 2 ras. (6) I.29 (7) But <KBC is right, hence <BCG is right. Thus the opposite <CGK & <GKB are also right. (7) Def. I.22; I.29; I.34 (8) Thus CGKB is a square described on CB. (9) For the same reasons HF described on HG = AC. (8) I.34 (10) Thus □HF = □AC (AC²) & □KC = □CB (CB²). (11) Since □AG = □GE & AG = AC(CB) b/c GC = CB. Thus also GE = AC(CB). (11) I.43; Def. II.1 (12) Thus □AG + □GE = 2(AC)(CB). But □HF = AC & □KC = CB . Hence, □HF + □CK + □AG + □GE = AC + CB + 2(AC)(CB). (12) C.N. 2 (13) But the four areas are = AB , thus AB = AC + CB + 2(AC)(CB). Therefore, etc. QED [Porism: Parallelograms about the diameter of a square are squares].46

45 Algebraic equivalent: (a + b) = a + b + 2ab 46 This porism was inserted by me because it seemed necessary for the propositions following II.4.

Proposition II.547

E: If SL is cut into equal and unequal segments, the product of the whole plus the SL between the points of section squared = the half [of the whole] squared. G: AB cut into equal segments @ C and unequal segments @ D. S: AD(DB) + CD = CB (1) Describe □CEFB on CB & join BE. (1) I.46; Post. 1 (2) Through D draw DG ll CE v BE; through H draw KM ll AB v EF; through A draw AK ll CL v BM. (2) I.31 (3) Thus, □CH = □HF. Add DM to each: □CH + □DM = □HF + □DM. Hence, □CM = □DF. (3) I.43; C.N. 2 (4) But □CM = □AL b/c AC = CB. Hence, □AL = □DF. (4) I.36; C.N. 1 (5) Add □CH to each: □AL + □CH = □DF + □CH. Hence, □AH = Γ NOP48 (b/c □AH = □AL + □CH & Γ NOP = □CH + □DF & □DF = □AL). (5) C. N. 2 (6) But □AH = AD(DB), b/c DH = DB.49 (6) I.29; Def. I.22; I.5; C.N. 1; I.6 (7) Thus Γ NOP = AD(DB). (7) C.N. 1 (8) Add □LG (= CD²) to each: Γ NOP + □LG = AD(DB) + □LG. Thus, Γ NOP + □LG = AD(DB) + CD . (8) See (6). (9) But Γ NOP + □LG = CB². Thus, AD(DB) + CD = CB . (9) C.N. 1 Therefore, etc. QED

47 Algebraic equivalent: ab + ((a +b)/2 – b) = ((a + b)/2) or (a + b)(a – b) = a + b depending on which segments one chooses for a and b. 48 I have not found a keyboard symbol for gnomons. Hence, Γ is my symbol for them. 49 Though this is proved in II.4, II.4 does not set out to prove this, hence it cannot be given as the reason. Hence, we must include all the reasons that DH = DB. Furthermore, it is an excellent exercise for the students.

Proposition II.650

E: If a SL be bisected & a SL be added to it, the product of the whole multiplied by the added SL plus the SL plus the half squared = the half plus the added SL squared. G: AB bisected @ C & BD added to it. S: AD(DB) + CB = CD (1) Describe □CEFD on CD & join DE. (1) I.46; Post. 1 (2) Through B draw BG ll EC or DF; through H draw KM ll AB or EF; through A draw AK ll CL or DM. (2) I.31 (3) Since AC = CB, □AL = □CH. (3) I.36 (4) But □CH = □HF, thus □AL = □HF. (4) I.43; C.N. 1 (5) Add CM to each: □AL + □CM = □HF + □CM. Hence, □AM = Γ NOP. (5) C.N. 2 (6) But □AM = AD(DB) b/c DM = DB. (6) I.29; Def. I.22; I.5; C.N.1; I.6; I.34. (7) Thus, Γ NOP = AD(DB). (7) C.N. 1 (8) Add □LG (=CB²) to each: Γ NOP + □LG = AD(DB) + □LG. Hence, AD(DB) + CB² = Γ NOP + □LG. (8) C.N. 2 (9) But Γ NOP + □LG = □CEFD on CD. Hence, AD(DB) + CB = CD . (9) C.N. 1 Therefore, etc. QED

50 Algebraic equivalent: (2a + b)b + a = (a + b) or (a + b)b + (a/2) = (a/2 + b)

Proposition II.751

E: If a SL is cut @ random, the whole squared plus one of the segments squared = two times the product of whole segment multiplied by the same segment plus the remaining segment squared. G: AB cut @ random @ C. S: AB + BC = 2(AB)(BC) + AC . (1) Describe □ADEB on AB & draw the figure. (1) I.46; Post. 1; I.31 (2) Since □AF = □GE, add □CF to each: □AF + □CF = □GE + □CF. Thus, □AF = □CE. (2) I.43; C.N. 2 (3) Hence, □AF + □CE = 2□AF. (3) P.O.M. (4) But □AF + □CE = Γ KLM + □CF; Thus, Γ KLM + □CE = 2□AF. (4) C.N. 1 (5) But 2(AB)(BC) = 2□AF b/c BF = BC. (5) I.29; Def. I.22; I.5; C.N. 1; I.6; I.34. (6) Thus, Γ KLM + □CF = 2(AB)(BC). (6) C. N. 1 (7) Add □DG (=AC²) to each: Γ KLM + □CF + □DG = 2(AB)(AC) + □DG. Thus, Γ KLM + □BG + □DG = 2(AB)(AC) + AC . (7) C.N. 2 (8) But Γ KLM + □BG + □DG = □ADEB + □CF (which are AB + BC ). (9) Thus, AB + BC = 2(AB)(BC) + AC . (9) C.N. 1 Therefore, etc. QED

51 Algebraic equivalent: (a + b) + b = 2ab + a

Proposition II.1152

E: To cut a SL so that the product of the whole multiplied by one of the segments = the remaining segment squared. G: AB. S: To cut AB: AB(HB) = AH 53 (find a point H) (1) Describe □ABDC on AB; Bisect AC @ E & join BE. (1) I.46; I.10; Post. 1 (2) Draw CA through to F & make EF = BE. (2) Post. 2; I.3 (3) Describe □FH on AF & draw GH through to K. (3) I.46; Post. 2 (4) More specific: AB is cut: AB(HB) = AH (5) Since AC is bisected @ E & FA is added to it, CF(FA) + AE = EF . (5) II.6 (6) But EF = BE, thus, CF(FA) + AE = BE . (6) P.O.M.; C.N. 1 (7) But AB + AE = BE . Thus, CF(FA) + AE = AB + AE . (7) Def. I.22; I.47; C.N.1 (8) Subtract AE from each: CF(FA) = AB . (8) C.N. 3 (9) Now CF(FA) = □FK b/c AF = FG & AB² = □AD. Hence, □FK = □AD. (9) C.N. 1 (10) Subtract □AK from each: □FK - □AK = □AD - □AK. Hence, □FH = □HD. (10) C.N. 3 (11) But □HD = AB(BH) b/c AB = BD & □FH = AH . (11) Def. I.22 (12) Thus, AB(BH) = AH . (12) C.N. 1 Therefore, etc. QEF

52 CF. VI.30. If we view this proposition in light of VI.30, we see that the line is cut in the extreme and mean ratio (i.e., the Golden Ratio). This proposition along with our knowledge of Book VI shows that we can construct the golden ratio on a finite straight line: Since AB = AH + BH, we get AH:BH::(AH + BH):AH. 53 Algebraic equivalent: solves the equation a(a – x) = x or x + ax = a with (a + b)b = a or ab = (a – b) .

Proposition II.1254

E: In obtuse-angled Δs, the side subtending the obtuse angle squared is > the sum of the sides containing the obtuse < squared by 2x the product of the side about the obtuse < & the SL cut off outside by the perpendicular towards the obtuse <. G: ΔABC w/ <BAC obtuse & draw BD from B ┴ CA extended to D. S: BC = BA + AC + 2(AC)(AD) [a = (1) Since CD has been cut @ random @ A, DC = CA + AD + 2(CA)(AD). (1) II.4 (2) Add DB to each: DC + DB = CA + AD + 2(CA)(AD) + DB (2) C.N. 2 (3) But CB = CD + DB & AB = AD + DB (3) I.47 (4) Thus CB = CA + AB + 2(CA)(AD) b/c CB = CD + DB . (4) C.N. 1 Therefore, etc. QED

Proposition II.1355

E: In acute-angled Δs the side subtending the acute angle squared is < the sum of the sides containing the acute angle squared by 2x the product of the segment containing the acute angle (upon which the perpendicular line falls) multiplied by the segment cut off within [the triangle] by the perpendicular towards the acute. G: Acute-angled ΔABC having <B acute & AD drawn from A ┴ BC. S: AC = CB + BA - 2(CB)(BD). (1) Since CB has been cut @ random @ D, CB + BD = 2(CB)(BD) + DC . (1) II.7 (2) Add DA to each: CB + BD + DA = 2(CB)(BD) + AD + DC . (2) C.N. 2 (3) But AB = BD + AD & AC = AD + DC . (3) I.47 (4) Hence, CB + BA = AC + 2(CB)(BD). (4) C. N. 1 (5) Thus, AC = CB + BA - 2(CB)(BD). (5) C. N. 3 Therefore, etc. QED

54 Algebraic equivalent: a² = b² + c² + 2b(c∙x/c)! or, importing trigonometric functions anachronistically, a² = b² + c² + 2b(c ∙ cos α) but, cos α = -cos θ thus a² = b² + c² - 2bc(cos θ) where θ = 180 degs. - α. Law of Cosines for obtuse triangles. 55 Algebraic equivalent: b = a + c - 2a(c ∙ cos θ). Law of Cosines for acute triangles. It is indeed, mutatis mutandis, the same law in II.12.

Proposition II.14

E: To construct a square = to a given rectilineal figure. G: Trapezoid A. S: To construct a square = to trapezoid A.56 (1) Construct □BD = trap. A. (1) I.45 (2) If BE = ED, then we are finished. If not, either BE > DE Or DE > BE. Let BE > DE. (2) Def. I.22 (3) Extend BE to F & make EF = ED. (3) Post. 2; I.3 (4) Bisect BF @ G. (4) I.10 (5) W/ centre G & distance GB (or GF) describe semicircle BHF. (5) Post. 3; Def. I.18 (6) Extend DE to H & join GH. (6) Post. 2; Post. 1 (7) Since BF is bisected @ G and divided unequally @ E, Seq. (BE)(EF) + EG = GF . (7) II.5 (8) But GF = GH, hence BE(EF) + GE = GH . (8) C.N. 1 (9) But HE + EG = GH . Hence, BE(EF) + GE = HE + EG . (9) I.47; C.N. 1 (10) Subtract GE from each: BE(EF) = HE . (10) C.N. 3 (11) But BE(EF) = □BD b/c EF = ED. Thus, □BD = HE² & □BD = trap. A. Hence, HE² = trap. A. (11) C.N.1; C.N. 1 Therefore, a square (on EH) has been described on HE = trap. A. QEF

56 Of interest is that the square we set out to construct is never explicitly constructed. It would seem that a step to describe a square on HE (I.46) is necessary unless one assumes that using I.47 means that we actually describe squares on the sides, but this does not follow Euclid’s typical path.

Book IV

Proposition IV.2

E: In a given ○ to inscribe a Δ =<ar w/ a given Δ. G: ○ABC & ΔDEF. S: To inscribe in ○ABC a Δ =<ar to ΔDEF. (1) Draw GH tangent to ○ABC @ A. (1) III.17 (2) On AH @ A make <HAC = <DEF & on AG @ A make <GAB = <DFE. (2) I.23 (3) Join BC. And since AH is tangent to ○ABC & AC is drawn from A across the circle, seq. <HAC = <ABC. (3) Post. 1; III.32 (4) But <HAC = <DEF, hence, <ABC = DEF. (4) C.N. 1 (5) For the same reason, <ACB = <DFE (i.e., since <GAB = <ACB & <GAB = <DFE, seq. <ACB = <DFE). (5) III.32; C.N. 1 (6) Thus <BAC = <EDF. (6) I.32 Therefore, in a given ○, we have inscribed a Δ =<ar to a given Δ. QEF

Proposition IV.3

E: About a given ○ to circumscribe a Δ =<ar w/ a given Δ. G: ○ABC & ΔDEF S: To circumscribe about the ○ABC a Δ =<ar w/ ΔDEF. (1) Extend EF in both directions to the points G, H. (1) Post. 2 (2) Take centre K of ○ABC & draw KB across ○ABC @ random. (2) III.1; Post. 1 (3) On KB & @ K make <BKA = <DEG & <BKC = <DFH. (3) I.23 (4) And through A, B, C draw LAM, MBN, NCL tangent to ○ABC. (4) III.17 (5) Since LM, MN, & NL are tangent to ○ABC @ A, B, C, & KA, KB, KC have been joined from centre E to the points A, B, C; thus, <s @ A, B, C are right. (5) III.18 (6) Since trap. AMBK = 4 ras. inasmuch as it divisible into 2 Δs & KAM, KBM are right, seq. <AKB + <AMB = 2ras. (6) C.N. 2; I.32; C.N. 3 (7) But <DEG + <DEF = 2ras. Hence, <AKB + <AMB = <DEG + <DEF.57 (7) I.13; C.N. 1 (8) But <AKB = < DEG, hence, <AMB = <DEF. (8) I.32; C.N. 3 (9) Similarly it can be proven that <LNB = <DFE.58 (9) I.32: C.N.3 (10) Thus <MLN = <EDF. (10) I.32; C.N. 3 (11) Thus ΔMLN (=<ar ΔDEF) has been circumscribed about ○ABC. Therefore, etc. QEF

57 In what respect do we say these angles are equal here? 58 The proof is as follows: Since trap. KBCN = 4 ras. & KCN, KBN are right, seq. <LNB + <BKC = 2ras. (I.32). But <DFH + <DFE = 2ras. (I.13). Hence, <LNB + <BKC = <DFH + <DFE. (C.N. 1). But <BKC = <DFH, thus <LNB = <DFE (I.32; C.N. 3).

Proposition IV.4

E: In a given Δ to inscribe a ○. G: ΔABC S: To inscribe a ○ in ΔABC (1) Bisect <ABC & <ACB w/ BD, CD. (1) I.9 (2) Let BD and CD meet @ D. Draw DE, DF, DG ┴ AB, BC, CA. (2) Post. 5; I.12 (3) Since <ABD = <CBD & <BED = <BFD, seq. ΔEBD # ΔFBD. (3) Post. 4; I.26 (AAS) (4) Hence, DE = DF. For the same reason, DG = DF. (4) Post. 4; I.26 (AAS) (5) Thus DE = DF = DG. (5) C.N. 1 (6) Hence, DE, DF, DG touch AB, BC, CA respectively b/c <s @ E, F, G are right. It will not cut them and hence, touches them & the circle will be inscribed in ΔABC. (6) III.16 (7) Inscribe ○FGE w/ centre D & distance DE v DG v DF. (7) Post. 3; Def. IV.5 Therefore in ΔABC, ○FGE has been inscribed. QEF

Proposition IV.5

E: About a given Δ to circumscribe a ○. G: ΔABC S: To circumscribe ΔABC with a circle. (1) Bisect AB, AC @ D, E and from D, E draw DE, EF ┴ AB, AC. (1) I.10; I.11 (2) They will meet either w/in ΔABC, on BC, v outside BC. (3) First let them meet w/in @ F & join FB, FC, FA. (3 Post. 1 (4) Since AD = DB, and DF is common, ΔDFB # ΔDFA Thus, AF = FB. (4 I.4 (SAS) (5) Similarly, we can prove CF = AF & FB = FC, hence, FA = FC = FB. (5) I.4 (SAS); C.N. 1 (6) Thus the ○ w/ centre F & distance FA v FC v FB, will pass through the remaining points and will circumscribe ΔABC. (6) Def. I.15; Def. IV.6 (7) Let it be circumscribed ○ABC. (7) Post. 3 (8) Next, let DF, EF meet on BC @ F & join AF.59 (8) I.10; I.11; Post. 1 (9) We can prove again that F is the centre of the ○ circumscribed about ΔABC. (9) Cf. above. (10) Again, let DF, EF meet outside ΔABC @ F & join AF, BF, CF. (10) I.10; I.11; Post. 1 (11) Since AD = DB & DF is common, ΔFDB # ΔFDA. Hence, AF = FB. (11) I.4 (SAS) (12) For the same reason, CF= AF, hence BF = CF. (12) I.4 (SAS); C.N. 1 (13) Thus the ○ w/ centre F & distance FA v FB v FC will pass through the remaining points and circumscribes ΔABC. (13) Def. I.15; Def. IV.6 Therefore about the given Δ a ○ has been circumscribed. QEF Corollary: When the centre falls w/in the Δ, the angle (BAC) is in a segment > than a semicircle & is < ra.; When the centre of the circle falls on BC, the angle (BAC) is in a semicircle & is = ra.; And when the centre of the circle falls w/out the triangle the angle (BAC) is in a segment < semicircle it is > ra. (III.31)

59 It is assumed that D, F still bisect AB, AC and are drawn at right angles to DF, FE & the proof would proceed in identical fashion to the first.

Proposition IV.6

E: In a given circle to inscribe a square. G: ○ABCD. S: To inscribe a square in ○ABCD (1) Draw two diameters AC ┴ BD & join AB, BC, CD, AD. (1) III.1; def. I.17; I.11; Post. 1 (2) Since BE = ED & EA is common, seq. ΔABE # ΔAED. Hence, AB = AD. (2) def. I.15; Post. 4; I.4 (SAS) (3) For the same reason, BC = AB & CD = AD. (3) def. I.15; Post. 4; I.4 (SAS) (4) Thus □ABCD is equilateral (BC = AB = CD = AD). (4) C.N. 1 (5) It is also right-angled. (6) Since BD is a diameter, seq. BAD is a semicircle, hence <BAD is right. (6) def. I.18; III.31 (7) For the same reason, <ABC, <BCD, & <CDA are right. (7) def. I.18; III.31 (8) Thus □ABCD is right-angled. But it was also proved to be =lateral. Hence, □ABCD is a square. (8) def. I.22 Therefore, in a given circle □ABCD has been inscribed. QEF

Proposition IV.7

E: About a given circle to circumscribe a square G: ○ABCD S: To circumscribe a square about ○ABCD. (1) Draw 2 diameters AC ┴ BD of ○ABCD. (1) def. I.17; I.12 (2) Through A, B, C, D, draw FG, GH, HK, KF tangent to ○ABCD. (2) III.16 Por. (3) Since FG is tangent & EA joined from the centre, seq. <s @ A are right. (3) III.18 (4) For the same reason, <s @ B,C,D are right. (4) III.18 (5) Since <AEB & <EBG are right, seq. GH ll AC. (5) I.28 (6) For the same reason, AC ll FK, hence GH ll FK. (6) I.28; I.30 (7) We can prove the same for GF ll HK ll BED. (8) Thus GK, GC, AK, FB, BK are □s. Hence, GF = HK & GH = FK. (8) I.34 (9) Since AC = BD & AC = GH = FK & BD = GF = HK, □FGHK is equilateral. (9) I.34 (10) It is also right-angled. Since GBEA is a parallelogram & <AEB is right, seq. <AGB is right. (10) I.34 (11) Similarly, we can show that <s H, K, F. are right. (11) I.34 (12) Thus □FGHK is right-angled. But it is also =lateral. Thus it is a square. (12) def. I.22 Therefore about a given circle a square has been circumscribed. QEF

Proposition IV.8

E: In a given square to inscribe a circle. G: □ABCD S: To inscribe a ○ in □ABCD (1) Bisect AD, AB in □ABCD @ E, F respectively. (1) I.10 (2) Through E draw EH ll AB v CD & through F draw FK ll AD v BC. (2) I.31 (3) Thus AK, KB, AH, HD, AG, GC, BG, GD are □s. Hence, In each figure the opposite sides are =. (3) I.34 (4) Since AD = AB & AE = ½AD, AF = ½AB, seq. AE = AF. (4) Def. I.22; POM; C.N. 1 (5) Thus the opposite sides are =. Hence, FG = GE. (5) I.34 (6) Similarly we can prove that each of SLs GH = FG & FG = GE. Thus GE = GF = GH = GK. (6) Def. I.22; POM; C.N. 1; I.34 (7) Thus the ○ w/ centre G & distance GE, GF, GH, v GK will pass through the remaining points. ` (7) def. I.15 (8) And it will touch the SLs AB, BC, CD, DA b/c <s E,F,K,H are right. (8) III.18 (9) For if the ○ cuts AB, BC, CD, DA the SL drawn ┴ to the diameter will fall w/in the ○, which is absurd. (9) III.16 (10) Thus the ○ will not cut AB, BC, CD, v DA. Hence, it will touch them & will have been inscribed in □ABCD. (11) Let it be inscribed as ○ABCD. (10) Post. 3 Therefore, in the given square a ○ has been inscribed. QEF

Proposition IV.9

E: About a given square to circumscribe a circle. G: □ABCD S: To circumscribe □ABCD w/ a ○. (1) Join AC, BD & let them cut one another @ E. (1) Post. 1 (2) Since DA = AB & AC is common, seq. DC = BC. Thus, <DAC = <BAC. (2) def. I.22; I.8 (3) Thus <DAB is bisected by AC. (4) Similarly we can prove that each of the <s ABC, BCD, CDA are bisected by AC, DB. (4) def. I.22; I.8 (5) Since <DAB = <ABC & <EAB = 1/2 <DAB & <EBA = ½<ABC. Thus <EAB = <EBA hence, EA = EB. (5) C.N. 1; I.6 (6) Similarly, we can prove EA = EC & EB = ED. (7) Hence, EA = EB = EC = ED. Thus ○ w/ centre E & distance EA v EB v EC v ED will pass through remaining points. (8) Let it be circumscribed ○ABCD. (8) Post. 3 Therefore about a given square a circle has been circumscribed. QEF

Proposition IV.10

E: To construct an isosceles Δ having each of the <s @ the base = 2x the remaining one. G: AB (1) Cut AB @ C: AB(BC) = CA . (1) II.11 (2) W/ centre A & distance AB describe ○BDE. (2) Post. 3 (3) Fit BD = AC < diameter of BDE. (3) IV.1 (4) Join AD, DC & about ΔACD circumscribe ○ACD. (4) Post. 1; IV.5 (5) Since AB(BC) = CA & AC = BD, AB(BC) = BD . (5) POM; C.N. 1 (6) Since B is taken outside ○ACD & from B, BA, BD fall on ○ACD: one cuts it & the other falls on it. And AB(BC) = BD , seq. BD is tangent to ○ACD. (6) III.37 (7) Since BD is tangent & DC is drawn across from D, <BDC = <DAC. (7) III.32 (8) Since <BDC = <DAC add CDA to each: <BDA + <CDA = <DAC + <CDA (<BDA = <DAC + <CDA) (8) C.N. 2 (9) But <BCD = <CDA + <DAC. Thus <BDA = <BCD. (9) I.32; C.N. 1 (10) But <BDA = <CBD since AD = AB. Hence, <DBA = <BCD (10) Def. I. 15; I.5; C.N. 1 (11) Thus <BDA = <DBA = <BCD. (11) C.N. 1 (12) And since <DBC = <BCD, BD = DC. (12) I.6 (13) But BD = CA, thus CA = CD. Hence, <CDA = <DAC. (13) C.N. 1; I.5 (14) Thus <CDA + <DAC = 2<DAC. (14) POM (15) But <BCD = <CDA + <DAC, hence <BCD = 2<DAC. (15) C.N. 1 (16) But <BCD = <BDA = <DBA = 2<DAB. (16) C.N. 1 Therefore ΔABD has been constructed such that the <s @ the base are each 2x the remaining <. QEF

Proposition IV.11

E: In a given ○ to inscribe an =lateral, =<ar pentagon. G: ○ABCDE S: To inscribe in ○ABCD an =lateral, =<ar pentagon. (1) Construct ΔFGH: the base <s @ G,H are each 2x < @ F. (1) IV.10 (2) Inscribe ΔACD =<ar ΔFGH in ○ABCDE: <CAD = <GFH; <ACD = <FGH; & <CDA = <FHG. (2) IV.2 (3) Thus <ACD & <CDA are each = 2<CAD. (3) POM (4) Bisect <ACD w/ CE & <CDA w/ DB. Then join AB, BC, DE, & EA. (4) I.9; Post. 1 (5) Since <ACD & <CDA = 2<CAD & are bisected by CE, DB, seq. <DAC = <ACE = <ECD = <CDB = <BDA. (5) POM (6) But = <s stand on = arcs. Hence, arcs AB = BC = CD = DE =EA. (6) III.26 (7) But = arcs are subtended by = chords. Hence, SLs AB = BC = CD = DE = EA. (7) III.29 (8) Thus pent. ABCDE is =lateral. (9) It is also =<ar. (10) Since arc AB = arc DE add arc BCD to each: arc ABCD = arc EDCB. (10) C.N. 2 (11) And <AED stands on arc ABCD & <BAE on arc EDCB Hence, <BAE = <AED. (11) III.27 (converse of III.26) (12) For the same reason, <ABC, <BCD, & <CDE are each = <BAE, <AED. (12) III.27; C.N. 2; C.N. 1 (13) Thus pent. ABCD is =<ar & =lateral. Therefore in a given ○ an =lateral. =<ar pentagon has been inscribed. QEF

Proposition IV.12 (In two parts)

E: About a given ○ to circumscribe an =lateral, =<ar pentagon. G: ○ABCDE S: To circumscribe an =lateral, =<ar pentagon about ○ABCDE Part I: (1) Let A, B, C, D, E be the angular points of a pent. Inscribed in ○ABCDE: arcs AB = BC = CD = DE = EA. (1) IV.11; III.26 (2) Through these <ar points draw GH, HK, KL, LM, & MG Tangent to ○ABCDE. (2) III.16 Por.; I.11 (3) Take centre F of ○ABCDE & join FB, FK, FC, FL, FD. (3) III.1; Post. 1 (4) Since KL is tangent @ C & FC has been joined from the Centre, to C, seq. FC ┴ KL. Hence, each < @ C is right. (4) III.18 (5) For the same reason, <s @ B, D are also right. (5) III.18 (6) Since <FCK is right, FK = FC + CK . (6) I.47 (7) For the same reason, FK = FB + BK , thus FC + CK = FB + BK . (7) I.47; C.N. 1 (8) But FC = FB , hence CK = BK . (8) def. I.15; POM; C.N. 3 (9) Thus CK = BK. (9) POM (10) Since FB = FC & FK is common [& BK = CK], <BFK = <KFC & <BKF = <FKC. (10) I.8 (SSS) (11) Thus <BFC = 2<KFC & <BKC = 2<FKC. (11) POM (12) For the same reason, <CFD = 2<CFL & <DLC = 2<FLC. (12) I.47; C.N. 1; Def. I .15: POM; C.N. 3; I.8; POM. Part II: (13) Since arc BC = CD, seq. <BFC = <CFD. (13) III.27 (14) And <BFC = 2<KFC & <CFD = 2<LFC. Hence, <KFC = <LFC. (14) POM (15) But <FCK = <FCL, <KFC = <LFC, & FC is common. Thus ΔFCK # ΔFLC. Hence, KC = CL & <FKC = <FLC. (15) Post. 4; I.26 (ASA) (16) Since KL = CL, KL = 2KC. For the same reason, HK = 2BK. (16) Steps (4) – (15) (17) And BK = KC, hence HK = KL. (17) C.N. 1 (18) Similarly each SL HG, GM, ML = HK, KL (18) Steps (4) – (17); C.N. 1 (19) Thus pent. GHKLM is =lateral. It is also =<ar. (20) Since <FKC = <FLC & <HKL = 2<FKC, <KLM = 2<FLC, Seq. <HKL = <KLM. (20) POM; C.N. 1 (21) Similarly we can prove <KHG = <HGM = <GML = HKL = <KLM. (21) POM; C.N. 1 (22) Thus pent GHKLM is =<ar & =lateral. Therefore, pent. GHKLM has been circumscribed about ○ABCDE. QEF

Proposition IV.13

E: In a given =lateral, =<ar pentagon to inscribed a circle. G: Regular Pentagon ABCDE S: To inscribe a circle in pentagon ABCDE. (1) Bisect <BCD & <CDE w/ CF, DF & from their point of intersection F, join FB, FA, FE. (1) I.9; Post. 1 (2) Since BC = CD & CF common, & <BCF = <DCF. Hence, ΔFBC # ΔFCD. Thus BF = DF & <CBF = <CDF. (2) I.4 (SAS) (3) Since <CDE = 2<CDF; <CDE = <ABC; & <CDF = <CBF, Seq. <ABC = 2<CBF. (3) C.N. 1 (4) Thus <ABF = <CBF. Hence, it is bisected by BF. (4) C.N. 3 (5) Similarly we can prove that <BAE, <AED are bisected by FA, FE respectively. (5) I.4; C.N. 1; C.N. 3 (6) Draw FG, FH, FK, FL, FM from F ┴ AB, BC, CD, DE, EA. (6) I.12 (7) Since <HCF = <KCF; <FHC = <FKC; & FC common, seq. ΔFHC # ΔFKC. Thus FH = FK. (7) I.26 (AAS) (8) Similarly we can show that each FL = FM = FG = FH = FK. (8) I.26; C.N. 1 (9) Thus the ○ described w/ centre F & distance FG v FH v FK v FL v FM will pass through the remaining points and be tangent to AB v BC v CD v DE v EA. (9) III.16 (10) Describe ○GHKLM. (10) Post. 3 Therefore, in a given =lateral, =<ar pent., a ○ has been inscribed. QEF

Proposition IV.15

E: In a given circle to inscribe a regular hexagon. G: ○ABCDEF S: To inscribe in ○ABCDEF a regular hexagon. Part I: (1) Draw diameter AD of ○ABCDEF. (1) def. I.17 (2) Take centre G of ○ABCDEF & w/ centre D & distance DG describe ○EGCH. (2) III.1; Post. 3 (3) Join EG, CG and extend them to B, F resp. (3) Post. 1; Post. 2 (4) Join AB, BC, CD, DE, EF, FA. Hex. ABCDEF is regular. (4) Post. 1 (5) Since GE = GD; DE = GD. Thus GE = DE & ΔEGD is =lateral. (5) def. I.15; C.N. 1 (6) Hence, <EGD = <GDF = <DEG. (6) I.5 (7) Since the interior <s of a Δ = 2ras., seq. <EGD = 1/3(2ras).60 (7) I.32; C.N. 3 (8) Similarly <DGC = 1/3(2ras.). (8) def. I.15;C.N. 1;I.5;I.32;C.N. 3 (9) Since CG on EB makes <EGC + <CGB = 2ras., seq. <CGB = 1/3(2ras.) [b/c <EGD + <CGD = 2/3(2ras.)]. (9) I.13 (10) Hence, <CGB = <DGC = <EGD & the <s vertical to them <BGA = <AGF = <FGE. (10) C.N. 1; I.15: (11) Thus the six <s are = to one another. (11) C.N. 1 (12) Hence, arcs AB = BC = CD = DE = EF = FA. (12) III.26 (13) Thus chords AB = BC = CD = DE = EF = FA & hexagon ABCDEF is =lateral. (13) III.29 Part II: (14) I say also that it is =<ar. Since arc FA = ED, add arc ABCD To each: FABCD = EDCBA. (14) C.N. 2 (15) And <FED stands on FABCD, while <AFE on EDCBA. <AFE = <DEF. (15) III.27 (16) Similarly it can be proved for the rest. (16) C.N. 2; III.27; C.N. 1 (17) Thus hex ABCDEF is regular. Therefore in a given circle a regular hex. Has been inscribed. QEF Porism: The side of the hexagon is = radius of the circle in which it is inscribed. Furthermore, if through the division on a circle we draw tangents to the circle, there will be a regular hexagon circumscribing the circle. Hence, we can inscribe a circle in a given hexagon and circumscribe a circle about it. Q.E.F.

60 Since <EGD + <GDF + <DEG = 2ras. Subtract <GDF + <DEG from each. <EGD - <GDF + <DEG = 2ras. - <GDF + <DEG. Thus <EGD = 1/3(2ras.).

Proposition VI.30

E: To cut a SL in the extreme and mean ratio. G: AB. S: To cut AB in the extreme and mean ratio. (1) Describe square BC on AB & apply to AC □CD = □BC & □AD exceeds the fig. (where □AD ~ □BC). (1) I.46; VI.29 (2) Now BC is a square hence, so is AD. (2) def. VI.1 (3) Since BC = CD take CE from each: BF = AD. (3) C.N. 3 (4) But since BF is also =<ar w/ AD the sides about the = <s are reciprocally proportional. Thus FE:ED::AE:EB. (4) VI.14 (5) But FE = AB & ED = AE. Hence, BA:AE::AE:EB. (5) def. I.22; C.N. 5 Therefore, AB has been cut in the extreme and mean ratio.61 QEF

61 This is the golden ratio (Euclid has ἄκρος καὶ μέσος λόγος) [golden section; divine section; divine proportion etc.] which is the whole is to the greater segment as that segment is to the other (lesser) segment. In other words, ϕ = a:b::(a + b):a (where a > b);ϕ = a/b = a + b/a (or a/a + b/a) hence ϕ = 1 + 1/ϕ. We can then turn it into a quadratic. Hence, ϕ = 1 + ϕ, then ϕ - ϕ – 1 = 0 (a = 1, b = -1, c = -1). Thus, from the quadratic formula we get ϕ = (1 + √5)/2 = 1.618033988 etc. We may also not that the Fibonacci numbers correspond to the GR. The sequence itself proceeds with the exception of the first two are the sum of the previous two numbers in the sequence. For instance, 1, 1, 2, 3, 5, 8, 13. 21, 34, 55, 89 etc. The ratio of the lead number in the sequence to the previous approaches the golden ratio. Have the class test this.