Ex-central Triangle
Subjects


  • GE 18 00 | - Outlines
  • GE 18 01 | - Diagram
  • GE 18 02 | - Definition of Ex-central triangle
  • GE 18 03 | - Ex-center : Concurrent of bisectors of triangle
  • GE 18 04 | - In-center of triangle ABC is ortho-center of triangle JKL
  • GE 18 05 | - The tangent from A to ex-circle
  • GE 18 06 | - Coordinate of ex-center
  • GE 18 07 | -
  • GE 18 08 | -
  • GE 18 09 | -
  • GE 18 10 | -
    • Answers


    GE 18 01. Diagram of ex-central triangle



    Diagram : Ex-central Triangle

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

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    GE 18 02. Definition of ex-central triangle

    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
    • 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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    GE 18 03. Prove that bisectors are concurrent

    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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    GE 18 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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    GE 18 05. The tangent from A to the ex-circle is AF = s

    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 GE 18

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    GE 18 06. Coordinate of ex-center

    Defintion : Relation with side AB
    • xh = s
    • yh = s*tan(A/2)

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    GE 18 07.

    Construction

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    GE 18 08.
    Construction

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    GE 18 09.

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    GE 18 10.

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    GE 18 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
    • Triangle ABC has 3 ex-circle
    Tnagents to ex-circle
    • From A to ex-circle J : AE = AF = s
    • From B to ex-circle J : BD = BF = s - b
    • From C to ex-circle J : CD = CE = s - c

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