Geometry Problem

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Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 913.

## Monday, August 12, 2013

### Problem 913: Right Triangle, Double Angle, Triple Angle, Metric Relations

Labels:
angle,
double angle,
metric relations,
right triangle,
triple

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http://img819.imageshack.us/img819/7592/s939.png

ReplyDeleteDraw F on CD such that CD=DE=e

Note that ED perpendicular to CF and triangle CEF is isosceles

In cyclic quadrilateral BCDE , angle(CED)= angle(CBD)= alpha= angle(DEF)

Triangle EAF similar to triangle CEF ………(case AA)

So triangle EAF is isosceles => d= B+2d

Peter, don't you need to prove that ED is perpendicular to CD? It's easy to do so but still it needs to be demonstrated, does it not?

DeleteAjit

DeleteYes,

By the result of problem 911, BCDE is a cyclic quadrilateral and hence angle CDE = angle CBE= 90

Peter.

My comment was quite unnecessary. I didn't see Problem #911 and went straight to 913. Hence the error.

DeleteExtend CD to F to form isoceles Tr. ECF with base < 90 - @ and so since < EAF = 2@ EA = AF and the result follows

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka