Thursday, January 7, 2010

Problem 415: Right Triangle, Cevian, Angles, Congruence

Proposed Problem
Click the figure below to see the complete problem 415 about Right Triangle, Cevian, Angles, Congruence.

Problem 415: Right Triangle, Cevian, Angles, Congruence.
See also:
Complete Problem 415
Level: High School, SAT Prep, College geometry

4 comments:

  1. E on DC
    Let
    m(DBE)=2a
    m(BDE)=m(BED)=90-a
    m(ABE)=90-a
    BD=BE
    AB=AE
    {AD=BA-DE
    EC=DC-DE
    BA=DC}
    AD=EC
    x=2a
    4a=90
    x=45

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  2. E is on AC such that BD=BE. <ABD=90-3a so <BED=<BDE=90-a=<ABE making AB=AE and triangles ABE and CDB congruent leading to AE=CB. AB=AE=CB so x=45.

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  3. Let BH be an altitude and let BE be the angle bisector of < HBC so that BH and BE trisect < 3@.

    Hence < ABH = x and < ABE = x + @ = < AEB so AB = AE = DC which in turn leads us to the conclusion that AD = EC

    Hence AB = BC and x = 45.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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