Monday, May 19, 2008

Geometry Problem 104



See complete Problem 104
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

5 comments:

  1. Let I be an incenter fo the trinagle ABD. Then
    angle of IBC = angle of IDC=90 degree and so the 4-points B,I,D,C on a circle, that is,
    the quadrilateral IDCB is a cycle.
    Hence, angle of BCD = alpha +beta and
    angle of 180- alpha -beta= 90+((angle of )/2).
    Furthermore,
    angle of DIC=90-alpha
    =(angle of A)/2 +(angle of IDA) and hence
    the points A, I, E, C are on the same line, that is,
    the points A, I, E, C are on the extended diagonal of the cicumcircle of the
    quadrilateral IDCB. and so
    x= angle of AEB=90+beta-alpha.

    2009.9.20
    From seoul, South Korea.
    Bae deok rak
    bdr@korea.com

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  2. This solution to problem 104 follows the same idea of Bae Deok Rak. I think that I’ve drawn it up in a clearer way.
    Let I be the incenter of triangle ABD. Then ang(IBC) = ang(IDC) = 90, the quadrilateral BCDI is cyclic, ang(BCI) = beta and ang(DCI) = alpha.
    We must prove that lines AC and IC are the same line.
    In triangle ADI, ang(AID) = 180 – A/2 – beta, and in triangle DIC, ang(DIC) = 90 – alpha. So, ang(AID) + ang(DIC) =
    = 270 – A/2 – alpha – beta (1).
    In triangle ABD, A = 180 – 2.alpha – 2.beta, or A/2 = 90 – alpha – beta. Therefore, in (1), we get ang(AID) + ang(DIC) =
    = 270 – (90 – alpha – beta) – alpha – beta = 180.
    This conclusion shows that A, I, E and C are collinear points.
    Finally, in triangle BCE, we have
    ang(AEB) = ang(DBC) + ang(BCI) = (90 – alpha) +beta = 90 + beta – alpha.

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  3. BC and DC are external bisectors of Tr ABD hence CA is the internal bisector and the result follows.

    sumith peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. How CA will be internal bisector pls prove

      Delete
    2. Extend AB to P, < PBC = 90 - alpha so BC is an external bisector of Tr. ABD

      Similarly DC is an external bisector

      Hence C is an excentre of Tr. ABD
      So CA must be an internal bisector

      Hope it’s clear now

      Delete