Fermat Point
The segments joining each F vertex of any triangle with the remote vertex of the equilateral triangle drawn D A externally on the L opposite side are concurrent, K O
C B
M
Proof : Circles K and L meet at points O and A. Since ADC = 240 and 1 1 ∠AOC = ADC , ∠AOC = 120 , Similarly, ∠AOB = AFB = 120, 2 2 ∴ ∠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. ) F LEMMA: (for Napoleon’s Theorem) D The segments joining each A vertex of any triangle L with the remote vertex K of the equilateral triangle drawn O externally on the opposite side are congruent, C B
M
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 . Napoleon’s Theorem
The circumcenters of the three equilateral triangles drawn externally on the sides of a given triangle determine an equilateral triangle. L A
K
B D O C
M
Proof : E Consider DAC . AK is two thirds the length of the altitude (or median). Using AC 3 the relationships in a 30-60-90 triangle = , Similarly AK 1
AF 3 AC AF in AFB, = . ∴ = . Since ∠KAC = ∠LAF = 30 , ∠CAL = ∠CAL , AL 1 AK AL
and ∠KAL = ∠CAF (addition), we have KAL CAF . Therefore, CF CA 3 = = . KL AK 1
DB 3 AE 3 DB AE CF Similarly, = and = . ∴ = = . But since KM 1 ML 1 KM ML KL
DB = AE = CF , we have KM = ML = KL or KML is equilateral. Menelaus’ Theorem
Q A Q A N N
P M P M R R
B L B L C C
Q A N
P M R
B L A set of 3 points located on each of the 3 sides. C
In any triangle, the Menelaus points are collinear if and only if AN / NB x BL / LC x CM / MA = -1 .
Proof : From similar triangles . AN AP BL QB CM RC = , = , and = NB QB LC CR MA AP By multiplication, we get A N AN BL CM ⋅ ⋅ = − 1 NB LC MA M
B L C Conversely, let the line containing N and L intersect AC at M ′ . Using the AN BL CM ′ converse just proved , we know that ⋅ ⋅ = − 1, But by hypothesis, NB LC M ′A
AN BL CM CM ′ CM ⋅ ⋅ = − 1 . ∴ = which indicates that M ′ = M . This proves NB LC MA M ′A MA collinearity. (Gerard) Desargues’s Theorem P
A
C B
M L N C' B'
are in “PERSPECTIVE” Copolar if AA', BB', and CC' are concurrent. Coaxial if intersection points of BC and A' B'C', CA and C'A', and AB and A'B' are collinear. 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.
Proof : Consider NBC to be a transversal of PB′C′ . Then by Menelaus’ Theorem PB NB′ CC′ ⋅ ⋅ = −1. Similarly, considering MBA to be a transversal of PB′A′ BB′ NC′ CP
PA A′M B′B we have ⋅ ⋅ = −1 . Now taking LAC to be a transversal of PA′C′ AA′ MB′ BP
PC C′L A′A we have ⋅ ⋅ = −1. Multiplying we get CC′ LA′ AP
B′N A′M C′L ⋅ ⋅ = −1 NC′ MB′ LA′
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. L
A'
A
M C C'
O B B'
N L
A'
A
M C C'
O B B'
N
Proof of Converse: Let N, M and L be collinear and consider AA′N and BB′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. Pascal’s Theorem
N
F E Y D M
C L A B X Z
The points of intersection of the opposite sides of an inscribed hexagon are collinear. ( Any conic ! ) 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:
XL ZE YD YN XA ZF ZM YC XB ⋅ ⋅ = − 1; ⋅ ⋅ = − 1 ; ⋅ ⋅ = − 1. LZ EY DX NX AZ FY MY CX BZ Taking the products of the three right members and the three left members, and rearranging the ratios,
XL ZM YN ZE ⋅ZF XB⋅ XA YC ⋅YD ⋅ ⋅ ⋅ ⋅ ⋅ = − 1 LZ MY NX ZB⋅ZA XC ⋅XD YE⋅YF Now each of the last three fractions in this product equals 1, we have XL ZM YN ⋅ ⋅ = − 1 and thus L, M, and N are collinear. LZ MY NX (Charles-Julien) Brianchon’s Theorem D E
d
e
c
F f O C
b a
A
B If a hexagon is circumscribed about a circle, the diagonals joining opposite vertices are concurrent.
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. (Robert) Simson Line
C
M
N B A
L P
The pedals drawn to the sides of a triangle from any point P on the circumcircle of the triangle are collinear
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. Euler Line
A
H
G O
B D A' C
P In any triangle, the centroid, orthocenter, and circumcenter are collinear.
Proof : Extend OG twice its distance to point H . (We will now show that H is the
orthocenter). AHG AOG , since AA′ is divided in a 2:1 ration by G , vertical angle at G , and GH = 2 ⋅OG by construction. This implies AH || OA′ 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. Lemma
C C
D
K P H H S S
A C' B A C' B
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.
Proof : Construct altitude AD and diameter KB intersecting AC at P . SC′ || KA and 1 SC′ = KA . Then CH || KA . Since ∠KCB = 90 and KC || AH ⇒ AHCK is a 2 parallelogram. ∴ CH = KA = 2SC′ (Karl) Feuerbach’s Theorem C
R D B' A'
E S (The Nine-Point Circle) A circle whose N center is the midpoint of the segment P H Q joining the orthocenter and the circumcenter, and whose radius is A F C' equal to half the radius of the B 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. Proof : Draw HS and C′R intersecting at N . (We will prove that N is the midpoint of 1 HS and that a circle with center N and a radius of CS will pass through 2 D, E, F, A′, B′, C′, P, Q, R .) 1 1. HR = HC = C′S . Also, HR || C′S . ∴ HC′SR is a parallelogram and 2
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 SC′ are equal and parallel. Therefore, CRC′S is a parallelogram 1 1 and C′R = SC (a circumradius). ∴ NR = C′R = SC . 2 2 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
C
6
5
4 7
Centroid Circumcenter
Nine-Point center 3 2 Orthocenter 8 9 A 1 B