## Friday, April 27, 2012

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

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

Click the figure below to see the problem 743 details. 1. A,B,F,C,E Concyclic
E,F,D,G Concyclic
Hence AngleEBC= AngleEFC = AngleEGD =28
x = 180 -90 - 28 = 62

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

Let ∠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

3. 4. Solution by Michael Tsourakakis from Greece

AC 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

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

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

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

Sumith Peiris
Moratuwa
Sri Lanka