Fermat Point

Proof : Circles K and L meet at points O and A. Since ADC  = 240 and

∠AOC  = 12 ADC , ∠AOC  = 120 , Similarly,

∠AOB =  12 AFB   =  120 ,

∴ ∠COB =  120 (because a complete revolution is 360 degrees).

Because CEB  = 240, ∠COB is an inscribed angle and point O must lie on circle M. Therefore, the three circles are concurrent, intersecting at point O.

Now, join point O with points A, B, C, D, E, and F. ∠DOA = ∠AOF  =  ∠FOB = 60 , and therefore DOB

. Similarly, COF

and AOE

. Therefore DB, AE, and CF are concurrent, intersecting at O ( which is also a point of intersection of circles K, L, and M. )

The segments joining each vertex of any triangle with the remote vertex of the equilateral triangle drawn externally on the opposite side are concurrent,

F

D

O

M

K

L

BC

A

F

D

O

M

K

L

BC

A

Proof : Since ∠DCA =  ∠ECB =  60, ∠DCB =  ∠ACE (addition). Also, DC =  AC and CB = CE , Therefore, DCB ≅ ACE  and DB =  AE . In a similar manner, we can show EBA ≅ CBF   ⇒  AE =  CF . Thus DB = AE =  CF .

LEMMA: (for Napoleon’s Theorem)

The segments joining each vertex of any triangle with the remote vertex of the equilateral triangle drawn externally on the opposite side are congruent,

E

D O

M

K

LA

C

B

Napoleon’s Theorem

Proof : Consider DAC . AK is two thirds the length of the altitude (or median). Using

the relationships in a 30-60-90 triangle ACAK  =   3 1 , Similarly

in

AFB, AFAL  =  3 1 . ∴ ACAK  =  

AFAL . Since ∠KAC  =  ∠LAF  =  30 , ∠CAL  =  ∠CAL ,

and ∠KAL  =  ∠CAF (addition), we have KAL  CAF . Therefore, CFKL  = 

CAAK =   3 1 .

Similarly, DBKM =   3 1 and AEML =   3 1 . ∴  DBKM  =  AEML =  CFKL . But since

DB = AE =  CF , we have KM  =  ML =  KL or KML is equilateral.

The circumcenters of the three equilateral triangles drawn externally on the sides of a given triangle determine an equilateral triangle.

LB

C

A

M

N

R

Q

P

L

R

Q

P

L

R

Q

P

LB

C

A

A

C

B

A

C

B

M

N

M

N

M

N

Menelaus’ Theorem

Proof :

From similar triangles .

 ANNB  = 

APQB , BL

LC  = QBCR , and CM

MA  =  RCAP

By multiplication, we get ANNB  ⋅ 

BLLC   ⋅ 

CMMA  =   − 1

Conversely, let the line containing N and L intersect AC at ′M . Using the

converse just proved , we know that ANNB  ⋅ 

BLLC   ⋅ 

C ′M′M A =   −  1, But by hypothesis,

ANNB  ⋅ 

BLLC   ⋅ 

CMMA  =   − 1 . ∴ C ′M

′M A =   CM

MA which indicates that ′M  = M . This proves

collinearity.

In any triangle, the Menelaus points are collinear if and only if AN / NB x BL / LC x CM / MA = -1 .

A set of 3 points located on each of the 3 sides.

C C'

O

L

N

M

A

A'

B

B '

LMN

A

BC

P

B 'C'

A '

(Gerard) Desargues’s Theorem

Proof : Consider NBC to be a transversal of P ′B ′C . Then by Menelaus’ Theorem PBB ′B

 ⋅ N ′BN ′C

⋅ C ′CCP  =  −1 . Similarly, considering MBA to be a transversal of P ′B ′A

we have PAA ′A

 ⋅  ′A MM ′B

⋅  ′B BBP  =  −1 . Now taking LAC to be a transversal of P ′A ′C

we have PCC ′C

 ⋅  ′C LL ′A

⋅  ′A AAP  =  −1 . Multiplying we get

′B NN ′C

 ⋅  ′A MM ′B

⋅  ′C LL ′A

 =  −1

If the triangles ABC and A’B’C’ are so situated that AA’, BB’, and CC’ are concurrent at P; and if BC and B’C’ meet at N, AB and A’B’ meet at M, and AC and A’C’ meet in L, then K, L, and M are collinear and conversely. Copolar triangles are coaxial, and conversely.

are in “PERSPECTIVE”

If the triangles ABC and A’B’C’ are in perspective, the points of intersection of the corresponding sides are collinear; and conversely, if the points of intersection of the corresponding sides are collinear, the triangles are in perspective.

Copolar if AA', BB', and CC' are concurrent. Coaxial if intersection points of BC and B'C', CA and C'A', and AB and A'B' are collinear.

C C'

O

L

N

M

A

A'

B

B '

Proof of Converse: Let N, M and L be collinear and consider A ′A N and B ′B M . These triangles are in perspective with L as center of perspective. Moreover, O, C and ′C are the points of intersection of their pairs of corresponding sides. Hence these points are collinear.

Y

X Z

L

M

N

C

D

EF

AB

Pascal’s Theorem

Proof : Consider XYZ whose vertices are the points of intersection of AB and CD , CD and EF , and EF and AB , respectively, and consider DE, FA and BC as transversals cutting the sides of XYZ . BY applying Menelaus’ Theorem we have the following relations:

XLLZ  ⋅ 

ZEEY   ⋅ 

YDDX  =  −  1; YNNX  ⋅ 

XAAZ   ⋅ 

ZFFY  =  − 1 ; ZMMY  ⋅ 

YCCX   ⋅ 

XBBZ  =   −  1.

Taking the products of the three right members and the three left members, and rearranging the ratios,

XLLZ  ⋅ 

ZMMY   ⋅ 

YNNX   ⋅ 

ZE ⋅ZFZB ⋅ZA  ⋅ 

XB⋅XAXC ⋅XD  ⋅ 

YC ⋅YDYE ⋅YF  =   −  1

Now each of the last three fractions in this product equals 1, we have XLLZ  ⋅ 

ZMMY   ⋅ 

YNNX    =  − 1 and thus L, M ,  and N   are collinear.

The points of intersection of the opposite sides of an inscribed hexagon are collinear. ( Any conic ! )

(Charles-Julien) Brianchon’s Theorem

Proof : Let ABCDEF be a hexagon circumscribed about a circle. The polars of the vertices A, B, C, D, E, F form the sides of the inscribed polygon whose sides are a, b, c, d, e,  f . Since a is a polar of A and d is a polar of D , then the point of intersection of a and d , call it L , is the pole of AD , Likewise, the point of intersection of b and e , call it M , is the pole of BE , And, the point of intersection of c and f , call it N , is the pole of CF , Therefore, by Pascal’s

Theorem,  L, M , N are collinear, hence their polars AD, BE, CF are concurrent.

If a hexagon is circumscribed about a circle, the diagonals joining opposite vertices are concurrent.

e

d

c

ba

f OF

A

B

C

DE

M

L

N

AB

C

P

(Robert) Simson Line

Proof :

Draw AP and PB . (We need to prove that ∠MNP + ∠PNL = st. ∠ ). SInce ∠AMP =  90  =  ∠ANP , A, P, N and M are concyclic. Also, since ∠PNB  = 90  = ∠BLP , P, L, B and N are concyclic. ∠MNP = ∠CBP , each being a supplement of ∠PAM . ∠CBP is the supplement of ∠PBL . But ∠PBL  =  ∠PNL since PLBN  is a cyclic quadrilateral. ∴∠MNP + ∠PNL  = st. ∠ and MNL   is a straight line.

The pedals drawn to the sides of a triangle from any point P on the circumcircle of the triangle are collinear

D

P

OG

A'

H

CB

A

Euler Line

Proof :

Extend OG twice its distance to point H . (We will now show that H is the orthocenter). AHG   AOG , since A ′A is divided in a 2:1 ration by G , vertical angle at G , and GH  =  2 ⋅OG by construction. This implies AH  || O ′A and thus perpendicular to BC . And since we could do the same construction from any vertex without moving the points O, G  and H and get the same result, we have shown that the three points lie on a single line.

In any triangle, the centroid, orthocenter, and circumcenter are collinear.

Lemma

Proof : Construct altitude AD and diameter KB intersecting AC at P . S ′C  || KA and

S ′C  = 12 KA  . Then CH  || KA . Since ∠KCB = 90 and  KC  || AH ⇒ AHCK is a

parallelogram. ∴ CH  = KA =  2S ′C

In any traingle, the segment of an altitude from a vertex to the orthocenter is twice the perpendicular from the circumcenter to the same side.

D

PK

H S

C'

C

A B

H

S

C'

C

A B

NS

P Q

R

H

D

E

F C'

A 'B '

C

AB

(The Nine-Point Circle) A circle whose center is the midpoint of the segment joining the orthocenter and the circumcenter, and whose radius is equal to half the radius of the circumcircle, passes through the feet of the altitudes, the midpoints of the sides, and the midpoints of the segments joining the vertices of the triangle to the orthocenter.

(Karl) Feuerbach’s Theorem

Proof : Draw HS and ′C R intersecting at N . (We will prove that N   is the midpoint of

HS and that a circle with center N and a radius of 12CS will pass through

D, E, F,  ′A ,  ′B ,  ′C , P, Q, R .)

1. HR =  12HC  =   ′C S . Also, HR ||  ′C S . ∴H ′C SR is a parallelogram and

HN  = NS and ′C N  =  NR . Therefore N is the midpoint of HS .

2. Since ∠ ′C FR = 90 , a circle with N as center and with radius NR will pass through F,   ′C ,  R .

3. CR  and S ′C   are equal and parallel. Therefore, CR ′C S is a parallelogram

and ′C R =  SC (a circumradius). ∴NR =  12 ′C R =  12SC .

We have proved that N is the midpoint of HS , that the circle with N as center and one half of SC (a circumradius) as a radius passes through a foot of an altitude, F , a midpoint of a side, ′C , and a midpoint of CH , R . Similarly the circle may be proved to pass through the other six points.

Extended Euler Line

Centro id

5

Nine-Point center

C i rcumcenter

2 8Orthocenter

6

3

1 9

74

C

A B