Monday, May 19, 2008

Geometry Problem 107

See complete Problem 107
Triangle, Angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Construct E on AC such that angles ABC=BEC, then angles CBD=EBD and we have DE/BE=DC/CB=AB/BC so triangles ABC~DEB giving triangle BAD isosceles thus x=pi-4a.


  3. Solution to problem 107.
    Take E on BC, such that AE is bisector of ang(BAC). Notice that ang(BAE) = ang(DBE) = alpha and both subtend the same segment DE. So ABED is cyclic, ang(BDE) = alpha, tr. BDE is isosceles and BE = DE.
    In triangle ABD, ang(BDC) = x + 2.alpha, so
    ang(CDE) = x + alpha.
    As AB = CD, BE = DE and
    ang(ABE) = ang(CDE) = x + alpha,
    triangles ABE and CDE are congruent, with
    ang(BAE) = ang(DCE) = alpha.
    Thus, in triangle ABC,
    (x + alpha) + 2.alpha + alpha = 180
    and finally x = 180 – 4.alpha.

  4. Let AE be the bisector of < BAC where E is on BC

    Tr. s ABE and CDE are congruent SAS so < EDC = < ABE implying that ABED is cyclic and < DBC = @

    So x = 180-4@

    Sumith Peiris
    Sri. Lanka

  5. Let E be on BC such that AE bisects < BAC

    Then ABED is cyclic and so Tr.s ABE and DEC are congruent SAS

    Hence AE = EC and so < ACE = @

    So x = 180-4@

    Sumith Peiris
    Sri Lanka

  6. Bring AE=//DC (AB=DC) then AECD is parallelogram so <BEA=<BAE=<CAE=a then AE is bisector angle BAC.Is <ECB=<DBC=a=<BAE so BECA is isosceles trapezoid or concyclic.
    Therefore <BCA=<BAE=a and <BDA=2a,so x=180-4a.