chapter 14 mathscape

531

Circle

geometry

This chapter at a glanceStage 5.1/5.2/5.3After completing this chapter, you should be able to:�� identify and name the parts of a circle�� use correct terminology associated with circle geometry�� use the chord properties to solve numerical problems�� use the chord properties to solve simple deductive problems, giving reasons�� use the angle properties to solve numerical problems�� use the angle properties to solve simple deductive problems, giving reasons�� use the properties of cyclic quadrilaterals to solve numerical problems�� use the properties of cyclic quadrilaterals to solve simple deductive problems, giving

reasons�� use the tangent properties to solve numerical problems�� use the tangent properties to solve simple deductive properties, giving reasons�� use the further circle properties involving chords, tangents and secants to solve

numerical problems�� use the further circle properties involving chords, tangents and secants to solve

simple deductive problems, giving reasons�� write formal deductive proofs involving circle properties

Cir

cle

geom

etry

14Mathscape 10 ext. - Ch14 Page 531 Friday, October 14, 2005 12:05 PM

Mathscape 10 Extens i on532

The following table shows the various parts of a circle and lists their definitions.

Circumference: the boundary of a circle. Semicircle: half the boundary of a circle.

Arc: part of the circumference. The smaller arc is called the minor arc and the larger arc is called the major arc.

Chord: an interval that joins two points on the circumference.

Diameter: a chord that passes through the centre.

Radius: an interval that joins the centre to a point on the circumference. It is half the length of the diameter.

Sector: the area bounded by an arc and two radii. The smaller sector is called the minor sector and the larger sector is called the major sector.

Segment: the area bounded by an arc and a chord. The smaller segment is called the minor segment and the larger segment is called the major segment.

Tangent: a line that touches a circle at one point.

Secant: a line that cuts a circle in two points.

14.1 Circle terminology

Mathscape 10 ext. - Ch14 Page 532 Saturday, September 24, 2005 12:46 PM

Chapter 14 : Circle geometry 533

Some other important terms are:

Concentric circles are two or more A cyclic quadrilateral is a quadrilateral incircles that have the same centre. which all of the vertices lie on a circle.

The points P, Q, R, S are concyclic points.

1 Name the parts of the circle shown on the diagrams below.a b c d

e f g h

i j k l

2 Name the following features that are shown on the diagram.a the interval OT b the interval ABc the line PS d the interval ATe the arc ADB f the line FGg the region OBET h the region ACT

3 On the given diagram, name all the:a radiib chords

O

Q

R

P

S

Exercise 14.1

O O

F AC P

T

SEB

G

DO

A

C

E

B

D

O



4 State the circle feature whose definition is given below.a Half of the circumference.b A section of the circumference.c An interval that joins two points on the circumference.d An interval that joins the centre of a circle to the circumference.e A line that touches a circle in exactly one point.f A chord that passes through the centre of a circle.g The area bounded by a chord and the circumference.h A line that cuts a circle in two points.i The area bounded by two radii and the circumference.

■ Consolidation

5 A circle has a radius of length 12 cm.a What is the length of the longest chord in the circle?b What name is given to this chord?

6 ∆JOK is drawn in a circle with centre O.a Explain why OJ = OK.b What kind of triangle is OJK?c Which angles are equal?

7 C and D are two points on a circle with centre O such that ∠COD = 60°. What kind of triangle is COD? Why?

8 Two concentric circles have diameters 22 cm and 14 cm. How far apart are the circles?

9 The circles shown touch internally at C. The larger circle has centre O. If the diameters of these circles are 20 cm and 7 cm, find OB.

10 Name the angle at the centre of this circle that is:a standing on the minor arc PQb subtended by the minor arc RS.

11 a Name the 3 angles at the circumference standing on the arc:i JK ii LM

b Name the 3 angles at the circumference standing on the arc:i JM ii LN

J

KO

O B C A

OS

R

P

Q

L

K

N

J

M

Mathscape 10 ext. - Ch14 Page 534 Tuesday, October 4, 2005 2:51 PM


12 a Name the 2 angles standing on the minor arc TW.b Name the angle standing on the major arc:

i TW ii TV iii SU

13 PA and PS are the intercepts of the secant drawn to the circle from P. Name the intercepts of the secant drawn froma G b X.

14 ∠ABC is an angle in a semi-circle, since the angle is subtended at the circumference by the diameter. Name two other angles in a semi-circle.

15 Is ∠PQR an angle in a semi-circle? Explain.

16 Find the size of each angle in the given diagram.a ∠ODEb ∠CODc ∠OCDd ∠CDE

17 a Name the angle at the centre, standing on the minor arc PR.b Name all angles at the circumference, standing on the

minor arc PR.

18 a Are the points K, L, M, N concyclic points?b What name is given to the quadrilateral KLMN?c Which angle is opposite ∠K?

V

U

S

T

W

E

A

N

S

X CR

G

P

E

A

B

D

CO

R

E

Q

P

FO

C

D

EO

30°

P

Q

S

O

R

T

N

L

M

K

Mathscape 10 ext. - Ch14 35 Saturday, September 24, 2005 12:46 PM


19 Is OXYZ a cyclic quadrilateral? Explain.

20 PQRS is a cyclic quadrilateral. If PQ is produced to T, name the exterior angle at Q.

21 Two intersecting circles with centres O and C have a common chord AB. What kind of quadrilateral is OACB if:a the circles are the same size? b the circles are different sizes?

■ Further applications

22 Draw a circle of radius 3 cm and draw any chord in the circle. Using a ruler and compasses, construct the perpendicular bisector of the chord. Does it pass through the centre of the circle?

23 Draw a circle of radius 3 cm and any two non-intersecting chords. Outline the constructions that could be used to locate the centre of the circle.

Data: O is the centre, OM ⊥ AB.Aim: To prove that OM bisects AB.Construction: Join OA and OB.Proof: In ∆s OAM, OBM

• ∠OMA = ∠OMB = 90° (OM ⊥ AB)• OA = OB (equal radii)• OM is a common side∴ ∆OAM ≡ ∆OBM (RHS)∴ AM = MB (matching sides of congruent ∆s)∴ OM bisects AB.

That is, the line drawn through the centre of a circle perpendicular to a chord bisects the chord.

Y

ZX

O

14.2 Chord properties of circles

The line drawn through the centre of a circle perpendicular to a chord bisects the chord.

A

O

BM



Data: O is the centre, M is the midpoint of AB.Aim: To prove that OM is perpendicular to AB.Construction: Join OA and OB.Proof: In ∆s OAM, OBM

• OA = OB (equal radii)• AM = MB (M is the midpoint of AB)• OM is a common side∴ ∆OAM ≡ ∆OBM (SSS)∴ ∠OMA = ∠OMB (matching ∠s of congruent ∆s)But, ∠OMA + ∠OMB = 180° (adjacent ∠s on a straight line)∴ ∠OMA = ∠OMB = 90°∴ OM ⊥ AB.

That is, the line drawn through the centre of a circle to the midpoint of a chord is perpendicular to the chord.

In the first proof above, it was proven that the perpendicular drawn through the centre O to a chord bisects the chord. Hence, this line is the perpendicular bisector of the chord. Therefore, the perpendicular bisector of the chord passes through the centre of the circle.

Data: O is the centre, AB = CD, OM ⊥ AB, ON ⊥ CD.Aim: i To prove that ∠AOB = ∠COD.

ii To prove that OM = ON.Construction: Join OA, OB, OC and OD.Proof: i • OA = OD (equal radii)

• OB = OC (equal radii)• AB = CD (given)∴ ∆AOB ≡ ∆DOC (SSS)∴ ∠AOB = ∠DOC (matching ∠s of congruent ∆s).

That is, equal chords of a circle subtend equal angles at the centre.

ii OM = ON (altitudes of congruent ∆s are equal).

The line drawn through the centre of a circle to the midpoint of a chord is perpendicular to the chord.

A

O

BM

The perpendicular bisector of a chord of a circle passes through the centre.

Equal chords of a circle subtend equal angles at the centre and are equidistant from the centre.

AO

B

M

C

D

N



Data: ABC is any triangle, O is the point of intersection of OX and OY, the perpendicular bisectors of ABand AC respectively.

Aim: To prove that the circle with centre O and radius OA also passes through B and C.

Construction: Join OA, OB, OC.Proof: In ∆s AOX and BOX

• AX = XB (OX bisects AB)• ∠AXO = ∠BXO = 90° (OX ⊥ AB)• OX is a common side∴ ∆AOX ≡ ∆BOX (SAS)∴ OA = OB (matching sides of congruent ∆s)

In ∆s AOY and COY• AY = YC (OY bisects AC)• ∠AYO = ∠CYO = 90° (OY ⊥ AC)• OY is a common side∴ ∆AOY ≡ ∆COY (SAS)∴ OA = OC (matching sides of congruent ∆s)Now, OA = OB and OA = OC, ∴ OA = OB = OC.

Hence, the circle with centre O and radius OA also passes through B and C. That is, given any three non-collinear points, the point of intersection of the perpendicular bisectors of any two sides formed by the three points is the centre of the circle through all three points.

Data: O, C are the centres, AB is a common chord.Aim: To prove that AB bisects OC at right angles.Construction: Join OA, OB, CA, CB.Proof: In ∆s OAC and OBC

• OA = OB (equal radii)• CA = CB (equal radii)• OC is a common side∴ ∆OAC ≡ ∆OBC (SSS)∴ ∠AOC = ∠BOC (matching ∠s of congruent ∆s)

Given any three non-collinear points, the point of intersection of the perpendicular bisectors of any two sides formed by the three points is the centre of the circle through all three points.

A

OB C

X Y

When two circles intersect, the line joining their centres bisects their common chord at right angles.

A

O

B

CM



In ∆s AOM and BOM• OA = OB (equal radii)• ∠AOC = ∠BOC (proven above)• OM is a common side∴ ∆AOM ≡ ∆BOM (SAS)∴ AM = MB (matching sides of congruent ∆s)∴ OM bisects AB.Also, ∠AMO = ∠BMO (matching ∠s of congruent ∆s).But, ∠AMO + ∠BMO = 180° (adjacent ∠s on a straight line).∴ ∠AMO = ∠BMO = 90°∴ OM ⊥ AB.

That is, when two circles intersect, the line joining their centres bisects their common chord at right angles.

Example 1a b

PQ = 12 cm. Find PM, OB = 15 cm and OM = 12 cm.giving reasons. Find AB, giving reasons.

Solutions

a PM = PQ (line through centre perpendicular to a chord bisects the chord)

∴ PM = × 12 cm

= 6 cmb i OB2 = MB2 + OM2 (Pythagoras’ theorem)

152 = MB2 + 122

MB2 = 81∴ MB = 9 cm

ii AB = 2MB (line through centre perpendicular to a chord bisects the chord)∴ AB = 2 × 9 cm

= 18 cm

EG +S

O

QMP

O B

M

A

12---

12---



Example 2a b

LM = NP and OX = 7 cm. AB = CD and ∠AOB = 40°.Find OY, giving reasons. Find ∠COD, giving reasons.

Solutionsa OY = OX (equal chords are equidistant from the centre)

∴ OY = 7 cmb ∠COD = ∠AOB (equal chords subtend equal angles at the centre)

∴ ∠COD = 40°

1 a b c

AB = CD and OP = 7 cm. OM = ON and PQ = 10 cm. OU = OV = 6 cm andFind OQ. Find RS. XY = 9 cm. Find YZ.

d e f

QR = RS and OA = 6.5 cm. OW = OX = 9 cm and GH = IJ = 20 cm andFind OB. KL = 25 cm. Find MN. OD = 8 cm. Find OC.

EG +S

O

N

M

7cm

P

YL

X

O

B

D

C

A

40°

Exercise 14.2

O

B

D

C

A

Q

P

O

S

R

P

M

N

Q

O

U

Y

ZV

X

O

R

B

A

S

Q

O

L

N

X

M

W

KO

C HI

G

J

D



2 Find the value of the pronumeral in each of these.a b c

3 a b c

AB = CD, ∠AOB = 110°, UV = 7 cm, YZ = 4 cm, PQ = QR = RS = ST = TP.∠COB = 65°. Find ∠BOD. WX = 6 cm. Find VW, Find ∠PQR.

XY, UZ.

4 a b c

PQ = 10 cm. Find PM. AM = 7 cm. Find AB. OM = ON, YZ = 22 cm.Find WX, WM.

d e f

EF = 15 cm, GQ = 7.5 cm, ∠ROS = ∠UOT, IJ = 20 cm, OM = ON,OP = 9 cm. Find GH, OQ. UT = 13 cm. Find XS. LX = 4 cm. Find XN.

O

52°8m

8m

x°

O

47° 47°

pm 11m

Oc °

O

D

B

A

C

O X

W

U

V

YZ

OR

QP

S

T

O

QPM

OB

A

M

OZ

W M X

Y

N

OH

F

E

P

G

Q

O

S

Y

T

U

RX

O JN

K

I

MX

L



■ Consolidation

5 a b c

OD = 13 cm, OM = 5 cm. PM = MQ = 20 cm, AB = 52 cm, OM = 10 cm.Find MD, CD. OM = 21 cm. Find OP. Find XY.

d e f

EF = 24 cm, FG = 30 cm. OM = 15 cm, YZ = 14 cm, AB = 80 cm, OD = 50 cm,Find OM, HM. ON = 24 cm. Find CD = 96 cm. Find

OZ, WX. OM, MN.g h i

FG = 12 cm, OG = 10 cm, OR = 39 cm, OQ = 25 cm, AB = 74 cm, BC = 70 cm.OH = 17 cm. Find GH. OM = 15 cm. Find QR, PR. Find OM.

6 Two circles with centres O and C intersect at A and B as shown. Let M be the point of intersection of the common chord and the line joining the centres of the circles.a If OC = 42 cm, OA = 20 cm and AB = 32 cm, find AC,

the radius of the larger circle.b If OA = 29 cm, AC = 35 cm and AB = 42 cm, find OC,

the distance between the centres.


7 A chord of length 18 cm is 12 cm from the centre of a circle. How far is a chord of length 10 cm from the centre? Answer in simplest surd form.

O

DMC

O

QMP

O

YMX

BA

O

EMF

GH

O X

MW

ZY N

O

BMA

DN

C

O

HMEGF O

RMP

Q

O

C

M

A B

CM

A

B

O



8 Two chords are drawn parallel to each other in a circle of radius 17 cm. The chords have lengths 16 cm and 30 cm. Find the distance between the chords if they lie:a on the same side of the centre b on opposite sides of the centre.

9 In the diagram shown, TU = 120 cm, VW = 105 cm and OM = 36 cm. Find SV.

10 Two circles with centres O and C intersect at P and Q. OC and PQ meet at M. If the circles have radii of 25 cm, 26 cm and the centres are 17 cm apart, find the length of the common chord PQ. [Hint: Let OM = x, PM = y.]

Data: O is the centre.Aim: To prove that ∠AOC = 2∠ABC.Construction: Join BO and produce BO to P.Proof: (In both figure 1 and figure 2)

In ∆AOB, let ∠OBA = α• OA = OB (equal radii)∴ ∆AOB is isosceles• ∠OAB = α (base ∠s of isosceles ∆, OA = OB)• ∠AOP = 2α (external ∠ of ∆ AOB)

OU

M

SV

T

W

Circumcircle symmetry

XYZ is an equilateral triangle of side length x cm, drawn in a circle centre O with radius r cm. Prove x2 = 3r2.

O

Z

X

Y x

r

TRY THIS

14.3 Angle properties of circles

The angle at the centre of a circle is twice the angle at the circumference, standing on the same arc.

O

C

figure 1

P

B

A



In ∆BOC, let ∠OBC = β• OB = OC (equal radii)∴ ∆BOC is isosceles• ∠OCB = β (base ∠s of isosceles ∆, OB = OC)• ∠POC = 2β (external ∠ of ∆BOC)

Now, in figure 1, ∠ABC = ∠OBA + ∠OBC= α + β

And, ∠AOC = ∠AOP + ∠POC= 2α + 2β= 2(α + β)= 2∠ABC

Also, in figure 2, ∠ABC = ∠OBC − ∠OBA= β − α

And, ∠AOC = ∠POC − ∠AOP= 2β − 2α= 2(β − α)= 2∠ABC

That is, the angle at the centre of a circle is twice the angle at the circumference, standing on the same arc.

Data: O is the centre.Aim: To prove that ∠ABC = ∠ADC.Construction: Join OA and OC.Proof: Let ∠ABC = α

• ∠AOC = 2α (angle at centre is twice angle at circumference)• ∠ADC = α (angle at centre is twice angle at circumference)∴ ∠ABC = ∠ADCThat is, angles at the circumference, standing on the same arc, are equal.

Data: AC is a diameter.Aim: To prove that ∠ABC = 90°.Construction: Join AB and BC.Proof: • ∠AOC = 180° (AC is a straight line)

• ∠ABC = 90° (angle at centre is twice angle at the circumference)

That is, the angle in a semi-circle is a right angle.

O

C

αP

B

A

β

figure 2

Angles at the circumference, standing on the same arc, are equal.

O

C

B D

A

The angle in a semi-circle is a right angle.

OC

B

A



Example 1Find the value of the pronumeral in each of these, giving reasons.

a b c

Solutionsa c = 32 (angles at circumference standing on the same arc are equal)b m = 50 (angle at the centre is twice the angle at the circumference)c k = 90 (angle in a semi-circle)

Example 2Find the value of all pronumerals in each of these, giving reasons.

a b

Solutionsa • a = 31 (alternate angles in parallel lines)

• b = 62 (angle at centre is twice the angle at the circumference)b • f = 90 (angle in a semi-circle is a right angle)

• g = 38 (angle sum of a triangle)• h = 38 (angles at circumference standing on the same arc are equal)


EG +S

c°32°

O

m°

25°

O

k°

EG +S

O

a°b°31°

O h°f °

52°

g°

Exercise 14.3

Oa°

60°

O

e°

104° O

y°



d e f

g h i

2 Find the value of each pronumeral.a b c

d e f

g h i


O

w°12°O q°

28°

Ob°

73°

O

d°86°

O

n°

74° O

x° 66°

m°40° z°

25° c°118°

O82°

p° q°O

140°

x° y°

O

55°

f °e°

O 42°v°

u° 37°k°j°

O28°

t°

r°

s°

O

n°

Oa° O

u°

v°



■ Consolidation

4 Find the value of all pronumerals in each of the following.a b c

d e f

g h i

j k l

m n o

b°

a° 50°

45°

m°

n° 25°O

t°

u°

53°

31°

h°

g°67°

O

x°

y°

O

z°44°

e°O

f°15°

s ° Or°

28°

c°O

d°

65°

w°O

v°

20°

O

j°

k°

48°O a°

b°

140°

62°y°

x° 57°

31° w°

z°

31°O

a°

b°36°

Os°r°

127°v°w°




5 Find the value of all pronumerals.a b c

d e f

6 Find values for a, b, c.

Data: O is the centre, ABCD is a cyclic quadrilateral.Aim: To prove that ∠ABC + ∠ADC = 180°.Construction: Join OA and OC.Proof: Let ∠ABC = α and ∠ADC = β.

• ∠AOC (reflex) = 2α (angle at centre is twice angle at circumference)

• ∠AOC (obtuse) = 2β (angle at centre is twice angle at circumference)

Now, 2α + 2β = 360° (angles at a point)∴ α + β = 180°

∴ ∠ABC + ∠ADC = 180°

That is, the opposite angles of a cyclic quadrilateral are supplementary.

27°O

w° 76°

O

q°65°

z°

58°

t°O

16°

63°

x°

15°

38°

68°

s°O

26°

82°

c°

O

73°

a°

b°25°

14.4 Cyclic quadrilaterals

The opposite angles of a cyclic quadrilateral are supplementary.

O2α

A

B

C

D

α

β

2β



Data: ABCD is a cyclic quadrilateral. AD is produced to E.Aim: To prove that ∠CDE = ∠ABC.Proof: Let ∠ABC = α.

• ∠ADC = 180° − α (opposite angles of a cyclicquadrilateral are supplementary)

• ∠CDE = 180° − (180° − α) (adjacent ∠s on= α a straight line)

∴ ∠CDE = ∠ABC

That is, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Example 1Find the value of each pronumeral, giving reasons.

a b

Solutionsa • p + 65 = 180 (opposite angles of a cyclic quadrilateral are supplementary)

∴ p = 115• q + 147 = 180 (opposite angles of a cyclic quadrilateral are supplementary)

∴ q = 33b u = 103 (exterior angle of a cyclic quadrilateral is equal to interior opposite angle).

Example 2Find the value of all pronumerals, giving reasons.

a b

Solutionsa • j = 48 (base angles of an isosceles ∆)

• k = 48 (exterior angle of a cyclic quadrilateral is equal to the interior opposite angle)b • m = 66 (opposite angles of a cyclic quadrilateral are supplementary)

• n = 132 (angle at centre is twice the angle at the circumference)

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

A

B

C

D E

180° – α

α

α

EG +S

q°

p°147°

65°u°

103°

EG +S

j°

48°

k°

n°

114°

m°

O




d e f

g h i


d e f

Exercise 14.4

50°

x°

115°

k° m°

p°

q°135°

52° a°

b°

96°68° v°

u°127°

104°

t°

t°

y°

2y°

3y°

75°

w°70°

e°53°

g°

48°

4k°

88°37°(u –10)°

112°

(5r + 27)°



■ Consolidation

3 a Is OABC a cyclic quadrilateral? Explain.b Are the opposite angles in OABC supplementary?c Find the reflex ∠AOC.d Find the value of x.

4 Find the value of the pronumerals in each of the following.a b c

d e f

5 Find the value of all pronumerals.a b c

d e f

O

A

B

C

x°

150°

O

m°

130°

n°

O

f °

164°

e°

Op° 59°

q°

O

d°

124°

c°

O

x°

72°

y°s°

135°

r°

g°

103°

h°O

w°

65°

v°41°

a°

72°

b°56°

r°

51°

s°

q°

j°

116°k°

t°62°

s°



g h i

6 Find the value of all pronumerals in each of the following.a b c

d e f

g h i

j k l

d°

54°

c°O

e°

23°

f°

n°

m°

O

y°z°x°

55°

74°a°

110° c°

b°41°

p°

r°

q°

65°

43°

Ov° 27°

u° w°

O

z°y°x°

75°

43°

a°

125°

b°

O

g°

h°

39°

O

92°

r°

s°p°

q°

116°

40°

a°34°

b°47°65°

O

d°16°

O

141° s°

O

d°

c°e°

17°



m n o


7 Find values for p and q.

8 ∠ABC = 2∠AOC. Find ∠ADC.

9 Find values for w, x, y, z.

z°23°

x°39°

y°

O

146°

m°

O118°

x°

y°

88°

96°p°

q° 113°

O

A

B

C

D

x°y°

w°

z°

114°

Angle tangle

O is the centre of this circle.

∠ABO = 5° and ∠OCB = 40°.

What is the value of x?

x°

O

A

CB

TRY THIS



Data: O is the centre, AB is a tangent to the circle at T.Aim: To prove that OT ⊥ AB.Proof: Let P be any point on the tangent AB, other than T.

Now, since P lies outside the circle, OP must be greater than the length of the radius OT, for all positions of P on AB.That is, the shortest distance between O and the tangent AB is OT. However, the shortest distance between a point and a line is the perpendicular distance. Therefore, OT ⊥ AB.That is, the tangent to a circle is perpendicular to the radius at the point of contact.

Data: O is the centre. PA and PB are tangents.Aim: To prove that PA = PB.Construction: Join OA, OB, OP.Proof: In ∆s OAP and OBP

• ∠OAP = ∠OBP = 90° (tangents PA,PB are perpendicular to radii OA, OB)

• OP is a common side• OA = OB (equal radii)∴ ∆OAP ≡ ∆OBP (RHS)∴ PA = PB (matching sides of congruent ∆s)That is, the two tangents drawn to a circle from an external point are equal in length.

Data: TP is a tangent, O is the centre.Aim: To prove that ∠ATP = ∠ABT.Construction: Draw in the diameter CT and the chord AC.Proof: Let ∠ATP = α

• ∠CTP = 90° (tangent is perpendicular to a radius)• ∠CTA = 90° − α (adjacent angles in a right angle)• ∠CAT = 90° (angle in a semi-circle)• ∠ACT = α (angle sum of ∆ACT is 180°)• ∠ABT = α (angles at circumference standing on

same arc are equal)∴ ∠ATP = ∠ABT

That is, the angle between a tangent and a chord drawn to the point of contact is equal to the angle in the alternate segment.

14.5 Tangent properties of circles

The tangent to a circle is perpendicular to the radius at the point of contact.

O

A T BP

The two tangents drawn to a circle from an external point are equal in length.

O

A

B

P

The angle between a tangent and a chord drawn to the point of contact is equal to the angle in the alternate segment.

O A

T

B

P

C

90° –

α

αα

α



Data: Two circles with centres O, C touch at T.Aim: To prove that O, T, C are collinear points.Construction: Construct the common tangent AB at T.Proof: (for figure 1):

• ∠ATO = 90° (tangent is perpendicular to a radius)• ∠ATC = 90° (tangent is perpendicular to a radius)• ∠ATO + ∠ATC = 90° + 90°

= 180°∴ O, T, C are collinear.

Proof: (for figure 2):• ∠ATO = 90° (tangent is perpendicular to a radius)• ∠ATC = 90° (tangent is perpendicular to a radius)• ∠ATO − ∠ATC = 90° − 90°

= 0°∴ O, T, C are collinear.

That is, when two circles touch (internally or externally), their centres and the point of contact are collinear.

Example 1Find the value of each pronumeral, giving reasons. PT and PS are tangents.

a b c

SolutionsSolutionsa a = 90 (tangent is perpendicular to a radius)b y = 6 (tangents drawn from an external point are equal)c • PS = PT (tangents drawn from an external point are equal)

• n = 65 (base angles of an isosceles triangle)

When two circles touch, their centres and the point of contact are collinear.

O

A

Bfigure 1

CT

O

A

B

CT

figure 2

EG +S

O

P

T

a°

P

T

y cm

6 cm

S

P

T

n°

65°

S



Example 2PT is a tangent. Find values for a and b, giving reasons.

Solutions• a = 44 (angle in the alternate segment)• b = 78 (angle in the alternate segment)

Example 3Find values for x, y, z, giving reasons.

Solutions• x = 75 (angle in the alternate segment)• y = 75 (base angles of an isosceles triangle)• z = 105 (opposite angles of a cyclic quadrilateral are supplementary)

1 In each of the following, PT is a tangent at A. Find the value of each pronumeral.a b c

2 PQ and PR are tangents. Find the value of all pronumerals in each of these.a b c

P T

b°

78° 44°

a°EG +S

y°

x°

z°75°

EG +S

Exercise 14.5

O

A TP

a°

O

A

T

P

y° 52°

O

A TP

e°

34°

Q

P

x cm

R12 cm QP

w°

R

65°

Q

P

g°

R

70°



■ Consolidation

3 a b

QP, QR, SR, ST are tangents. QP = 7 cm BA, BC, DC, DE are tangents. BA = 6 cmand QS = 18 cm. Find ST. and DE = 8 cm. Find BD.

c d

UV, VW, UW are tangents. VZ = 5 cm, EF, FG, GH, EH are tangents. EK = 7 cm,UY = 8 cm and VW = 14 cm. Find the FK = 5 cm, FG = 8 cm and the perimeterperimeter of ∆UVW. of EFGH is 48 cm. Find HM.

4 In each of the following, PQ is a tangent at T.a b

OT = 12 cm and TQ = 35 cm. Find OQ. OQ = 26 cm and OT = 10 cm. Find TQ.

Q

R

q° P

53°

d Q

R

t°P 46° Ov°

u°

e

Q

R

e°P

113°O

15 cm

f cm

f

Q SR

P T

B DC

A E

U WY

Z X

V

E

G

F

NM

H

LK

O

P T Q

O

P T Q



c d

PT = 36 cm and OP = 39 cm. Find PR. RS = 28 cm and PT = 48 cm. Find PS.

5 LM is a tangent at T in each of the following. Find the value of all pronumerals.a b c

d e f

g h i

O

P

T

Q

R

OP

T

Q

R S

50°

LT

N

K

M

67°

x°

L

T

Q

M

49°

P

52°

L

T

F

M

41°E

68°p°

q°

LT

D

M

44°

C 82°

h°g°

L

T

X

M

17°

Y

108°u°v°

L

T

J

M

64°

I

78°f° e°

L

T

G

M

114°c°

H

L T

S

M

z°

R

12°

L

T

V

Ms°

U



6 PT is a tangent in each of the following. Find the value of all pronumerals.a b c

d e f

g h i

7 Find the value of the pronumeral in each of these. PT, PS, QR are tangents.a b c

T

F

b°

E

P

O

53°

a° T

Y

n°

X

P

O

132°

m°

T

L

e°

M

P

74°

f°

N

T

q°

C

P

35°

p°

DE

T

y°

P

12°

w°z°

x°O

T

P81°

s°

t°

r°O

G

H

h°

g°

15°28°

XY

Z

P

T

b°

d°

76°

123°

a°c°

B

A

C

P T

y°

z°

77°

x°

L

M

P

T

b°55°

O

24°

T

D

E

P

f°71°

T

B

A

P c°47°

TY

X P

119°

Z



d e f

j k l

t°

47°T

L

P

M

O m° 76°T R

PK

Q k°

85°T

HP

F

G33°

E

r°

78°

T

S

P

A

g

x°

68° R

S

T

P

h

Sy°

24°

D

PT

C

46°

i

v°

W

PT

X70°

O

32°

j°

W

P

T

V

113°U

25° e°

K

P

TJ

24°I

55°

80°

s°

CP

TB

A

62°

D

E

19°

m

a°

P

T S

O

U

74°

n

d°P

T

Y

Z

X

W

98°15°

o




8 A circle is inscribed in a right-angled triangle as shown. BQ = 5 cm and QC = 7 cm. Find the length of AC.

9

PT and PS are tangents. Find the value of x.

10 PQ is a tangent at T. Find the value of x.

Data: Two chords AB and CD intersect at P.Aim: To prove that AP . PB = CP . PDConstruction: Join AC and BD.Proof: In ∆s APC and DPB

• ∠CAP = ∠BDP (angles at circumference standing on same arc are equal)

• ∠APC = ∠DPB (vertically opposite angles)∴ ∆APC ||| ∆DPB (equiangular)

∴ (matching sides of similar ∆s are in proportion)

∴ AP . PB = CP . PDThat is, the products of the intercepts of two intersecting chords of a circle are equal.

P

C

B

Q

A

R

5 cm

7 cm

T

C

P

S

A

B

15°

O

D

100°

x°

T

X

P

WV

U

Qx°

Length of a median

ABD is a right-angled triangle and C is the midpointof BD. If AB = 6 cm and AD = 8 cm, find the exact length of AC.

B

A

C D

TRY THIS

14.6 Further circle properties

The product of the intercepts of two intersecting chords of a circle are equal.

BC

D

A

P

APPD-------- CP

PB--------=



Data: The secants PA and PC cut the circle at B and Drespectively.

Aim: To prove that PA . PB = PC . PDConstruction: Join AD and BC.Proof: In ∆s ADP and CBP

• ∠BAD = ∠BCD (angles at circumference standing on same arc are equal)• ∠APD = ∠CPB (common angle)∴ ∆ADP ||| ∆CBP (equiangular)


∴ PA . PB = PC . PD

That is, the products of the intercepts of two intersecting secants to a circle from an external point are equal.

Data: PT is a tangent, PA is a secant, which cuts the circle at B.Aim: To prove that PT 2 = PA . PBConstruction: Join AT and BT.Proof: In ∆s BPT and TPA

• ∠PTB = ∠TAB (angle in the alternate segment)• ∠TPB = ∠TPA (common angle)∴ ∆BPT ||| ∆TPA (equiangular)


∴ PT 2 = PA . PB

That is, the square of a tangent to a circle from an external point is equal to the product of the intercepts of any secants from the point.

Example 1Find the value of the pronumeral in each of the following, giving reasons. PT is a tangent in (c).

a b c

The products of the intercepts of two intersecting secants to a circle from an external point are equal.

B

C

D

A

P

PAPC-------- PD

PB--------=

The square of a tangent to a circle from an external point is equal to the product of the intercepts of a secant from the point.

B

A

T

P

PTPA------- PB

PT-------=

EG +S

B

D

A

X

C

m3

4

12

B

D

A

PC

a

3

2

7B

T

A

P

u

5

15



Solutionsa AX . XB = CX . XD (products of intercepts of intersecting chords)

∴ 4 × m = 3 × 124m = 36

∴ m = 9b PA . PB = PC . PD (products of intercepts of intersecting secants)

2(a + 2) = 3(3 + 7)2a + 4 = 30

2a = 26∴ a = 13

c PT 2 = PA . PB (square of tangent is equal to the product of the intercepts of the secant)u2 = 5(5 + 15)u2 = 100

∴ u = 10 (u > 0)

Example 2Find the value of t.

Solutiont(t + 2) = 4(4 + 8)t2 + 2t = 48

t2 + 2t − 48 = 0(t + 8)(t − 6) = 0

∴ t = −8, 6But t > 0, ∴ t = 6.


2 Find the value of each pronumeral.a b

t

4

8

2

EG +S

Exercise 14.6

m

4

8

5y

6

8

3

u

512

10

p

5

3

2j

5

3

4



c d

3 PT is a tangent in each diagram. Find the value of each pronumeral.a b

c d

■ Consolidation

4 Find the value of each pronumeral. PT is a tangent in d and e.a b c

d e

c

66

4

u

15

5

4

R

2

6

T Pa

Q

R

4

5

T

Pt

Q

R

4

8

T

P

n

Q

R

9

T

PQ

w

12

A

18

E

Bx

y

C

F

D

159

6U

16

Y

q

W

X

106

Z

Vp8

L

12

Hm

G

K

24

4 I

J

n

10

T

P

f

10

e

6

8

B

D

C

A

TP

r

16

s 9

5

B

D

A

C



5 O and Z are centres, PA is a tangent to the circle with centre O. PT is a common tangent to both circles.AP = 24 cm and PB = 18 cm.a How long is PT? Why?b Find the length of BC.

6

PT is a tangent. Find values for a and b.

7 Find the value of each pronumeral. PT is a tangent in a and b.a b

c d


8 In the diagram, PQ ⊥ RS, RS = 16 cm and MQ = 4 cm. Find the length of theradius.

T

P

18 cm

O

24 cmA

Z

C

B

T

P

4

R

S

Qa

b

3

5

26

R

P

6

Q

5

T

n

UP 3V

2

T

a

X

A

5 B

2

C

t

3

Y

M

3.5

6.5

c

2

E F

L

N

M

Q

O

SR

P



9 PT is a tangent, PR = 4 cm and PT = 8 cm. Find the length of the radius:a by using Pythagoras’ theoremb by using a circle property

The emphasis in this exercise is on the justification of each step in an argument. Full reasons must therefore be given in all questions. O is the centre of the circle, unless otherwise stated.

Example 1Find the value of the pronumeral in each of the following, giving reasons.a b

PT is a tangent.

Solutionsa OA = OC (equal radii)

∴ ∆OAC is isosceles∠OCA = 34° (base ∠s of an isosceles ∆, OA = OC)• ∠AOC = 112° (∠sum of ∆AOC)• ∠ABC = 56° (∠ at centre is twice ∠ at circumference)∴ a = 56

b • ∠TAC = 53° (∠ in the alternate segment)• ∠ACT = 53° (alternate ∠s, AC || TP)• ∠ATC = 74° (∠ sum of ∆ACT)• ∠ABT = 106° (opposite ∠s of a cyclic quad. are supplementary)∴ x = 106

Q

O

T

R

P

Deductive proofs involving circle properties

14.7

EG +S

C

O

B

A

a°

34°

C

B

T

x°

53°

A

P



Example 2ABCD is a cyclic quadrilateral. E and F are points on AD andBC respectively such that AB || EF. Prove that C, D, E, F areconcyclic points.

SolutionLet ∠ABC = α• ∠ADC = 180° − α (opposite ∠s of a cyclic quadrilateral are supplementary)• ∠EFC = α (corresponding ∠s, AB || EF)Now, ∠ADC + ∠EFC = 180° − α + α

= 180°∴ CDEF is a cyclic quadrilateral∴ C, D, E, F are concyclic points.

Example 3US bisects ∠RST. Prove that UQ bisects ∠PQT.

SolutionLet ∠TSU = α• ∠USR = α (US bisects ∠RST)• ∠TQU = α (∠s at circumference standing on

same arc are equal)• ∠PQU = α (external ∠ of cyclic quadrilateral QRSU is equal to interior opposite ∠)∴ ∠TQU = ∠PQU∴ UQ bisects ∠PQT

1 Find the value of each pronumeral, giving reasons. PT and PS are tangents.a b c

C

B

D

A

F

E

180° – α

α

α

EG +S

P

Q

S

U

R

T

α

α

α

α

EG +S

Exercise 14.7

O

CA

B

x°

54°

O

C A

B

e°

43°

D

O

C

A Bp°

D

12°



d e f

g h i

j k l

m n o

S

T

g° P

76°

Tu°

P

56°

O

D

c°P

74°69°A

B

D

m°C

112°A

B

E

F

G

O

a°

T

37°

AP

O B

A

Mr°

51°

O

B

44°A

k°

C

DE

O

B

126°

Af°

C

D

O

S

28°P

y°R

T

A

w°

B

T QP

52° A

n°

B

D

E

C38°

77°

A

z°

BD

Q

C

45°

T

P

86°



p q r

s t u

v w x

■ Consolidation

2 LM is a common tangent. LM, FG and EH meet at T, the point of contact of the circles.a Explain why ∠FTL = ∠MTG.b Hence, prove that EF || GH.

As°

B

D C

117°

S

q° PQ

T

78°C

n°

P

A

T

B

Q

0

26°

C

D

A

E

B

O

96°

v°

C

D

A

E

B

72°

r°

140° C

TA

Q

Bc°

21°

P

C

T

A

D

B

j°

79°

P

138°

C

T

A

QB

t°70°

P QC is also a tangent

36°

C

T

A

Q

B

u°

40°P

116°

T

M

H

G

F

E

L



3 OM ⊥ AB, ON ⊥ BC and BM = BN. Prove that OM = ON.

4 The circles shown are concentric and OM ⊥ PQ. Prove that PR = SQ.

5 In the diagram shown, AB || CD.Prove that ∆APB is isosceles.

6 Two circles with centres O and C touch internally at A.Prove that M is the midpoint of AB.

In the diagram shown, PQ is a tangent and PQ || RS.Prove that TU bisects RS.

8 Two chords AB and CD meet at P. PA = PC.a Prove that ∆DPB is isosceles.b Hence, prove that AB = CD.c Prove that AC || DB.

C

M

A

N

B

O

QP M

O

R S

D

P

C

B

A

M

OC

B

A

U

O

S

P

RV

TQ

7

B

P

D

A C



9 In the diagram, PQ || RS.Prove that ∠QTS = ∠PSR.

10 Two circles intersect at P and Q. PA and PB are diameters.Prove that A, Q, B are collinear.

11 From an external point P, two secants are drawn to a circle as shown. PA = PC. Prove that AB = CD.

12 In the diagram, TW = TX and ∠XOZ = ∠XTZ. Prove that ∠XOZ = ∠XTZ.

13 Two circles intersect at X and Y.AB and CD are straight lines.Prove that AC || BD.

T

P

R

Q

S

B

P

A Q

B

CA

P

D

Y

X

W

Z

O

T

Y

BX

DC

A



14 In the diagram, ∠CBD = ∠EFG.a Prove that ∠CBD = ∠EBA.b Hence, prove that PD = PE.

15 PQ is a tangent and AB = BC.a Prove that AB bisects ∠CAQ.b Prove that DB bisects ∠ADC.

16 PT is a tangent and RT bisects ∠STQ. Prove that PR = PT.

17 Two circles intersect at B and D.AB and DC are tangents and AB || DC.Prove that BC || AD.

18 PT and SQ are tangents. Prove that ∠TQS = ∠PRT.

F

B

P

D

C

A

EG

B

DC

A QP

Q

S

T P

R

B

D

A

C

T

S

Q

R

P

O



19 PT is a tangent. AT bisects ∠CTP and TC ⊥ AB.Prove that AB is a diameter.


20 PT is a tangent and PT = TN.a Prove that PM = MT.b Hence, prove that ∆MOT is equilateral.

21 AB is a tangent to both circles and PT is a common tangent. Prove that A, B, T are concyclic points.

22 PT is a tangent and CP bisects ∠BPT. Prove that ∆TCD is isosceles.

B

AC

PT

O

M

N

P

T

T

AA P B

T

APB

C

D



FO

CU

S O

N W

OR

KIN

G M

AT

HE

MA

TIC

AL

LY

FO

CU

S

ON

W

OR

KIN

G

MA

TH

EM

AT

IC

AL

LY

AN INTERESTING PROOF OF THE SINE RULE USINGCIRCLE GEOMETRY

Introduction

Circle geometry provides a very useful way to deepen your understanding of important trigonometric results. In the diagram above a triangle ABC has been drawn inside a circle centre O. In this focus on working mathematically we will use two circle theorems to prove the sine rule.

In the process we will discover an interesting link between the sine rule and circle geometry.

FO C U S O N WO R K I N G MA T H E M A T I C A L L Y0 F O C U S O N W 0 R K I N G M A T H E M A T I C A L L Y

a

A

c

B

figure 1

C

b

O

sin A sin B sin Ca = =b c

asinA----------- b

sinB----------- c

sinC-----------= =

Mathscape 10 ext. - Ch14 Page 574 Friday, October 14, 2005 10:18 AM


FO

CU

S O

N W

OR

KIN

G M

AT

HE

MA

TIC

AL

LY

FO

CU

S

ON

W

OR

KIN

G

MA

TH

EM

AT

IC

AL

LY

L E A R N I N G A C T I V I T I E S

1 Copy figure 1 into your book. If you can, construct the diagram using Geometer’s Sketchpad. This will enable you to drag the triangle around and get a better idea of why the proof works no matter where A, B and C are on the circle. Instructions for this are set out below.

2 We now need to draw some construction lines. Join B to O and produce the line to meet the circle at D. Now join DC. Label the radius of the circle R. Your figure will now look like this:

3 What special name is given to the straight line BD passing through O?

4 Look at the shaded triangle BDC. What angle in this triangle is equal to angle A in triangle BAC? Why? What circle theorem did you use? Mark the equal angles on your diagram.

5 What do you notice about the size of angle BCD? Why? What circle theorem did you use? Mark the size of the angle on your diagram.

6 Write down an expression for sinD in terms of a and R.

7 Hence show that and that .

22

a

A

B

C

O

D

R

R

figure 2

sinAa

2R-------= a

sinA----------- 2R=



FO

CU

S O

N W

OR

KIN

G M

AT

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MA

TIC

AL

LY

FO

CU

S

ON

W

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TH

EM

AT

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AL

LY

8 Repeat the argument, this time showing that, similarly, . It is a good idea to

draw a new diagram. Decide what diameter you will need to draw. Discuss this in class.

9 Can you see that the method will work for ? Discuss this in class. Complete the proof.

10 Complete this statement: in any given triangle ABC, each ratio in the sine rule

represents …

C H A L L E N G E A C T I V I T I E S

1 In the diagram on page 575, a = 10 cm and A = 60°. Calculate the diameter of the circumcircle of the triangle ABC (leave the answer in surd form).

2 From the result of question 1, calculate the length of DC (leave the answer in surd form).

3 If angle ACD is 45° show that angle ABC is 75°. Give reasons for each step of your argument.

4 Calculate b, the length of AC, and hence find the area of triangle ABC (answer to 2 decimal places).

Instructions for using Geometer’s Sketchpad

1 Drawing figure 1.

a Select the Square Grid from the Graph menu. Then hide the axes and reference points by selecting them and choosing Hide Objects from the Display menu.

b Use the Compass (circle) tool to draw the circle of radius 5 units. Select the point at the end of the radius and hide it. Choose Hide Point from the Display menu.

c Select the circle and then choose Point on Circle from the Construct menu. Repeat to create the points A, B and C on the circle. (relabel the centre and other points as necessary by double clicking on the label). Drag A, B and C around the circle to suitable positions as shown. Then select A, B and C and choose Segments from the Constructmenu to construct the triangle.

d Select each side and use the Text tool to label the length of the sides a, b and c.

2 Drawing figure 2.

a Select B and then O (in that order) and choose Line from the Construct menu.

b Select the circle and the line and choose Intersections from the Construct menu. Hide the intersection point which appears next to B and label the other intersection point D.

c Now select and hide the line. As you drag B around the circle you will see that D moves in the same way. It is always at the end of the diameter.

bsinB----------- 2R=

csinC-----------

asinA----------- b

sinB----------- c

sinC-----------= =

88



FO

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G M

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TIC

AL

LY

d Now select B, O and D and choose Segments from the Construct menu. B, O and D are all in a straight line, so BD is a diameter. Now join D to C. Select and label the radii R.

e Finally select the points D, B and C and choose Triangle Interior from the Constructmenu. Choose a colour from the range in the Display menu.

f Drag the points A, B and C around the circle to see the effect on the diagram. What geometrical properties remain constant? Can you see that the result is true no matter where A, B and C are on the circle?

L E T ’S C O M M U N I C A T E

In no more than half a page, summarise what new mathematics you have learned from this activity. Use a diagram to illustrate.

R E F L E C T I N G

Did it surprise you that the sine rule is so closely connected to the diameter of the circumcircle of the triangle? Why is trigonometry so closely linked to geometry?

EE

%%

1 Draw a diagram to clearly illustrate the meaning of each of the following:a An angle at the centre of a circleb An angle at the circumference of a

circlec An angle in a semicircled An angle between a tangent and a

chorde An angle in the alternate segment

2 Why is it so much easier to draw the angles above rather than describe them in words? Are there times when the English language actually appears to be a hindrance in communicating mathematical ideas?

3 Do you think it is possible for mathematicians who speak different languages, such as English and Chinese, to communicate mathematically? Why?



CH A P T E R RE V I E W

CH

AP

TE

R R

EV

IEW

O is the centre of the circle in all questions, unless otherwise specified.

1

Name the following parts of the circle.a the interval OEb the interval CDc the interval CFd the line ABe the line PTf the region EODGg the region CFHh the curved interval EGD

2 a Explain whyOA = OB.

b What kind of triangle is OAB?

c Which angles are equal?

3 What are concentric circles?4 Is OPQR a cyclic

quadrilateral? Explain.

5 a

b

c

d

e

f

E

A

G

DC

B

HF

P

T

O

O

BA

O

R

Q

P

O

A

BC

D42°

AB = CD.Find ∠COD.

O

P

Q

S

R

4 cm

∠POQ = ∠ROS.Find RS.

O

Y

W

Z

X

N

M3 cm

WX = YZ.Find ON.

O

G M HGM = 5 cm.Find GH.

O M

S

R

OR = 13 cm,OM = 5 cm.Find RS.

O

M

Z

N

W

YX

OM = ON,WX = 14 cm.Find NZ.




CH

AP

TE

R R

EV

IEW

6 Find the value of each pronumeral.a

b

c

7 Find the value of all pronumerals.a

b

c

d

8 Find the value of all pronumerals. PT and PS are tangents.a

b

c

C

T

A

D

B

j°

79°

138°

O

p°

c°

74°

O

110°125°

v°

u°

f °e°

136°113°

(2g + 12)° 66°

3h°100°

45°

6j°

(7k –18)°

O

k°

T P

S

T

P 52°y°

z cm

9 cm

59°

p°63°

q°




CH

AP

TE

R R

EV

IEW

9 Find the value of each pronumeral. PT is a tangent.a

b

c

d

e

f

10 Find the value of all pronumerals, giving reasons. PT and PS are tangents.a

b

X

AB

C

D

x

93

5

I

K

M

J

N

L

p

q12 8

9

16

R

Q

PT n

15

5

4

6 t

T

P

Q

R

C

E

D

B

A

u

10

86

6

35

Wk

YV

X

Z

A

B

PT

O17°

n°

82°

e°

OC

BA




CH

AP

TE

R R

EV

IEW

c

d

e

f

11 a

Prove that ∆RST is isosceles.b

• PT is a tangent• PT = PRProve that TR bisects ∠QTS.

c

i Prove that ∠AOC = 2∠OAB.ii Find ∠AOC.

d

• PT, PS are tangents• TS bisects ∠QTPi Prove that TQ || PS.ii Prove that ∆QST is isosceles.

A

B

PT

Og°

57°

T

P

S

A m°48°

77°

u°

T

P

AO

B

C

D

A

O

B

65° r°111°

Q

SP

R

T

• PQ || RS

T

PQ R S

A C

O

B

D• O is the centre• OABC is a rhombus

TP

S

Q


chapter 14 mathscape

Documents

circle terminologymathscape

chord properties

simple deductive properties

numerical problems

tangent properties

simple deductive problems

angle properties

minor arc