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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 743 details.
A,B,F,C,E ConcyclicE,F,D,G ConcyclicHence AngleEBC= AngleEFC = AngleEGD =28x = 180 -90 - 28 = 62
http://img259.imageshack.us/img259/43/problem743.png Let ∠CDE= α Note that ∠AGE= ∠CDE= αWe 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:https://docs.google.com/open?id=0B6XXCq92fLJJUk5oRWdKSTJJbEU
Solution by Michael Tsourakakis from GreeceAC 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
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 PeirisMoratuwaSri Lanka