Ex-central Triangle
Subjects

  • Q01 | - Diagram
  • Q02 | - Definition of Ex-central triangle
  • Q03 | - The proof of the formation of an ex-center
  • Q04 | - In-center of triangle ABC is ortho-center of triangle JKL
  • Q05 | - The tangent from A to ex-circle
  • Q06 | - The coordinate of ex-center
  • Q07 | - The angles of ex-central triangle JKL
  • Q08 | - The sides of ex-central triangle JKL
  • Q09 | - The area of ex-central triangle
  • Q10 | - The circum-radius of ex-central triangle
    • Answers


    Q01. Diagram of ex-central triangle



    Diagram :Ex-central Triangle

    Diagram and locus of ex-center
    • Study Subject GE 03 02 and GE 10 04
    • How to prove ? See program 13 03 in MD2002

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    Q02. Definition
    Construction
    • 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
    • Similarly triangle ABC has other ex-centers K and L
    • 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
    • 3. Prove that
      • JA perpendicular to KL
      • KB perpendicular to LJ
      • LC perpendicular to JK
    • 4. Prove that ortho-center of triangle JKL is coincided with in-center of ABC
    • 5. Ex-circle will tangent the sides of triangle ABC
    • 6. Prove that Tangen AF = AE = s where the tangent touch circle at E and F

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    Q03.
  • Proof : Incenter of ABC and orthocenter JKL are coincided

    Question
    • Let BJ and CJ are the bisectors
    • Prove that AJ is also a bisector and passing point J
    proof
    • Let JD is distance to line BC
    • Let JE is distance to line CA
    • Let JF is the distance to line AB
    • Since BJ is bisector, hence JD = JF
    • Since CJ is bisector, hence JD = JE
    • Hence JE = JF and then AJ is bector of angle A
    • Hence they are concurrent

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    Q04. 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
    Proof
    • 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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    Q05. 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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    Q06. 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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    Q07. 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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    Q08. 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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    Q09. The sides 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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    Q10. 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
    Answer

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