Geometry Problem

Click the figure below to see the complete problem 481 about Problem 481: Triangle, Altitude, Cevian, Perpendicular, Angle.

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Complete Problem 481

Level: High School, SAT Prep, College geometry

## Monday, July 26, 2010

### Problem 481: Triangle, Altitude, Cevian, Perpendicular, Angle

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Let (XYZ) denote angle XYZ

ReplyDeleteNote that quadrilateral ABFE is cyclic . (AEB)=(AFB)=90

We have (BAF)=(BEF) ( both angle face the same arc BF)

Angle alpha = (ABF)- (EBF) and angle beta= (EBG)-(EBF)

But ( ABF)= (EBG) ( both angle complementary to (BAF) and ( BEF) )

So alpha=beta

Peter Tran

I think its still easier...

ReplyDeleteAEFB cyclic implies that <ABE = <AFE=90-<BFG=90-(90-<FBG)=<FBG

as dessired.

Let BG and AF meet at X

ReplyDeleteThen < ABE = < AFE = < XFG = < FBX

Sumith Peiris

Moratuwa

Sri Lanka

ABFE is cyclic ==> <AFE = <FBG and we conclude that <ABE = <CBG.

ReplyDelete