Proposed Problem

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

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Complete Problem 415

Level: High School, SAT Prep, College geometry

## Thursday, January 7, 2010

### Problem 415: Right Triangle, Cevian, Angles, Congruence

Labels:
angle,
cevian,
congruence,
right triangle

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E on DC

ReplyDeleteLet

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

by sine rule on BAD triangle and BDC triangle

ReplyDeleteE 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.

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

ReplyDeleteHence < 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