Monday, May 19, 2008

Geometry Problem 106



See complete Problem 106
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

3 comments:

  1. Construct E on BC such that AE bisects angle BAC, then AE=CE implies triangles ABE and CDE are congruent, thus ABED is cyclic and angle DBC=a, therefore angle ABD=pi-4a.

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  2. Or
    extend CA such BA=AE; m(ABE)=m(BEA)=a
    BE=BC & DC=AE so AB=BD
    x=180-4a

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  3. The symmetrical point D about the BC is E, then AB = DC =EC and BD=DE,<BCE=a=<ECB, but <BAC=<ECA=2a and AB=CE then ABEC is isosceles trapezoid (ΒΕ//ΑC ),with <CBE=<ACB=a=<ECB so BE=EC=DC=BD . Then <DBC=<DCB=a, so <BDA=<DBC+<DCB=2a.Therefore x=180-2a-2a=180-4a.

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