Geometry Problem

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

Click the figure below to see the problem 743 details.

## Friday, April 27, 2012

### Problem 743: Trapezoid, Triangle, Perpendicular, Parallel, Angle

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A,B,F,C,E Concyclic

ReplyDeleteE,F,D,G Concyclic

Hence AngleEBC= AngleEFC = AngleEGD =28

x = 180 -90 - 28 = 62

http://img259.imageshack.us/img259/43/problem743.png

ReplyDeleteLet ∠CDE= α

Note that ∠AGE= ∠CDE= α

We have CE/ED=AE/GE= tan( α)

Triangle BAE similar to triangle DEG… ( Case SAS)

So ∠EDG= ∠ABE= 90-28= 62

The solution is uploaded to the foowing link:

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJUk5oRWdKSTJJbEU

Solution by Michael Tsourakakis from Greece

ReplyDeleteAC and DG meet at K. the triangle AGD, GE is height , DF is height .So C is orthocenter of triangle AGD so AK is height. SO the quadrangle ECKD is inscribable .so angles KDE and ACE are equal. but angle ACE is equal to angle EBA=62.So x=62

http://www.mathematica.gr/forum/viewtopic.php?f=22&t=27475&p=134337

ReplyDeleteABCE is a rectangle, so < AEB = 28 = < AFB since ABFC is cyclic.

ReplyDeleteSo ABFE is cyclic and so < ABE = 62 = < AFE = x since DEFG is cyclic.

Sumith Peiris

Moratuwa

Sri Lanka