Geometry Problem

Click the figure below to see the complete problem 541 about Right Triangle, Altitude, Angle Bisector, Congruence.

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## Thursday, November 25, 2010

### Problem 541: Right Triangle, Altitude, Angle Bisector, Congruence

Labels:
altitude,
angle bisector,
congruence,
right triangle

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draw EP perpendicular to AC

ReplyDelete▲ADF ~ ▲AEP => AP/EP = AD/DF

▲ABE ~ ▲AEP => AB/BE = AP/EP

AF bisector => AB/BF = AD/DF

=>

AB/BE = AB/BF

=>

BE = BF

Consider right triangles ABE, AFD

ReplyDeleteComplements of the equal angles BAE,FAD are equal

So Angle BEF = Angle AFD

= Angle BFE (vertically opposite)

etc

Triangle ABE ~ Triangle ADF by AA Test of similarity

ReplyDelete.....(angle FAD=angle BAF{given}, angle ABC=angle ADB=90{given})

By c.a.s.t angle AFD=angle FEB

angle AFD=angle BFE ....(vertically opposite angles)

Hence, angle BFE=angle FEB.

Therefore BF=BE....(Isosceles Angle Theorem)

The complements of both < BEA and AFD are A/2 and the result follows

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Problem 541

ReplyDeleteIs<BEF=<ECA+<EAC=90-<BAE=90-<A/2, but <BFE=<AFD=90-<FAD=90-<A/2=

Therefore BE=BF.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE