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Draw a large triangle ABC
Draw bisector of angle A
Draw bisector of external angle of angle B
Draw bisector of external angle of angle C
The bisectors are concurrent at point J which is the ex-ccenter
Ex-center J is between AB and AC
Similarly triangle ABC has
- Ex-center K is between BC and CA
- Ex-center L is bewteen CA abd AB
Join J, K, L to form triangle JKL which is called excentral triangle

Questions

1. Prove that the bisectors are concurrent
2. Prove that
- K, A, L are colinear
- L, B, J are colinear
- J, C, K are colinear
3. Prove that
- JA perpendicular to KL
- KB perpendicular to LJ
- LC perpendicular to JK

Properties

1. Ortho-center of triangle JKL is coincided with in-center of ABC
2. Ex-circle will tangent the sides of triangle ABC
3. Tangen AF = AE = s where the tangent touch circle at E and F

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TR 19 03. Prove that bisectors are concurren

Construction

Let BJ and CJ are the bisectors and intersect at J
Prove that AJ is a bisector and also passing point J

proof

Let JD is distance to line BC and D is on BC
Let JE is distance to line CA and E is on CA
Let JF is distance to line AB and F is on AB
By bisector of angle thoery
- Since BJ is bisector, hence JD = JF
- Since CJ is bisector, hence JD = JE
By bisector of angle thoery
- Since JE = JF
- Hence AJ is also bisector of angle A
Hence they are concurrent

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TR 19 04. In-center of triangle ABC is ortho-center of triangle IKL
Triangle ABC is pedal triangle of ex-central triangle JKL

J, C, K are colinear and LC perpendicular to JK
K, A, L are colinear and JA perpendicular to KL
L, B, J are colinear and KB perpendicular to LJ
Hence triangle ABC is pedal triangle of triangle JKL

Prove L,A,K are colinear

Produce CA to P
LA is bisector of angle BAP : Angle BAL = (Angle BAP)/2
JA is bisector of angle BAC : Angle JAB = (Angle BAC)/2
Hence angle JAL = (angle BAL) + (angle JAB) = (BAC + BAP)/2 = 90 degrees
Similarly angle JAK = 90
Hnece L,A,K are colinear

Proof JA perpendicular to LK

Produce CA to P
Angle PAB is external angle of angle A
Hence angle CAB + angle PAB = 180 degrees
Angle LAB + angle BAJ = (angle CAB + angle PAB)/2 = 90
Hence JA perpendicular to KL
Similarly KB perpendicular to LJ
Similarly LC perpendicular to JK
Hence KL, LJ, JK are concurrent at the in-center of traingle ABC
Hence ortho-center of triangle JKL coincides with in-center of triangle ABC
Note to remember
ex-center J is between AB and AC
ex-center K is between BC and BA
ex-center L is between CA and CB

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TR 19 05. The tangent from A to the ex-circle

Ex-circle is between AB abd BC

Let circle touch BC at D, AC at E and AB at F
AF + AE = AB + BF + AC + CE
By tangent rule, BF = BD and CE = CD
Hence AF + AE = AB + AC + BD + CD
Hence 2*AF = AB + BC + CA = 2*s
Hence AF = s

Tangent from B to ex-circle

Ex-circle is between AB abd BC
Tangents BF = BD = AF - AB
Hence BF = s - c

Tangent from C to ex-circle

Ex-circle is between AB abd BC
Tangents CE = CD = AE - CA
Hence BF = s - b

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TR 19 06. Coordinate of ex-center

Ex-radius r1 = AF*tan(A/2) = s*tan(A/2)
Let AB be in horizontal direction, then x = s and y = r1
The ex-center is (s, r1) if AB is along x-axis

Why we assume AB is in horizontal direction

It is easily locate the center on oxy coordinate
Otherwise we to use the rotation method to locate the center on oxy coordinate

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TR 19 07. The angles of ex-central triangle

Construction

Ex-center between AB abd BC is J
Ex-center between BC abd CA is K
Ex-center between CA abd AB is L

Find angles (See pedal triangle)

Since triangle ABC is pedal triangle of triangle JKL
Hence angle BAC = 180 - 2*(angle KJL)
Hence Angle KJL = 90 - (angle BAC)/2 = 90 - A/2 (Angle KJL oppsite angle A)
Similarly angle JKL = 90 - B/2 (Angle JKL oppsite angle B)
Similarly angle KLJ = 90 - C/2 (Angle KLJ oppsite angle C)

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TR 19 08. The sides of ex-central triangle
Construction

Ex-center between AB abd BC is J
Ex-center between BC abd CA is K
Ex-center between CA abd AB is L

Find sides

Since triangle ABC is pedal triangle of triangle JKL
Hence a = BC = (KL)*cos(LJK)
Hence KL = a/cos(LJK) = a/cos(90 - A/2) = a/sin(A/2)
By sinelaw a = 2*R*sin(A) = 4*R*sin(A/2)*cos(A/2)
Hence KL = 4*R*cos(A/2)
Similarly LK = 4*R*cos(B/2). (Note B on JL)
Similarly JL = 4*R*cos(C/2). (Note C on KJ)

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TR 19 09. The area of ex-central triangle

Area = (LJ)*(JK)*sin(KJL)/2
Area = (4*R*cos(A/2))*(4*R*cos(C/2))*sin(90 - A/2))/2
Area = 8*(R^2)*cos(A/2)*cos(B/2)*cos(C/2)
Where R is circum center of triangle ABC

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TR 19 10. The circum-radius of ex-central triangle

Circum-radius of ex-central angle
= KL/(2*sin(KJL)
= (4*R*cos(A/2))/(2*sin(90 - A/2))
= 2*R
Where R is circum center of triangle ABC

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TR 19 00. Outlines

Properties of ex-central triangle

J, C, K are colinear and LC perpendicualr to JK
K, A, L are colinear and LC perpendicualr to KL
L, B, J are colinear and LC perpendicualr to LJ
Triangle ABC is pedal triangle of ex-central triangle JKL
Incenter of triangle ABC coincdes with ortho-center of triangle JKL

Tnagents to ex-circle

From A to ex-circle J : AE = AF = S

Properties from trigonometry

Angle of JKL
- Angle KJL = 90 - A/2 (Angle KJL oppsite angle A)
- Angle JKL = 90 - B/2 (Angle JKL oppsite angle B)
- Angle KLJ = 90 - C/2 (Angle KLJ oppsite angle C)
Sides of JKL
- KJ = 4*R*cos(A/2). (Note A on LK)
- LK = 4*R*cos(B/2). (Note B on JL)
- JL = 4*R*cos(C/2). (Note C on KJ)
Coordinate of J (s, s*tan(A/2))
Area of JKL = 8*(R^2)*cos(A/2)*cos(B/2)*cos(C/2)
Circum-radius of JKL = 2*(circum-radius of ABC) = 2*R

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