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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 698.
Let AB and DC intersect at P. Since m(APD)=60 and m(APD)+m(BEC)=180 PBEC is cyclic quadrilateral. PE is angle bisector of BAC. So, m(BEC)=m(EPB)=30
http://img840.imageshack.us/img840/4838/problem698.pngExtend AB and CD to F ( See picture)Note that angle(A)+ Angle (D)=120 and angle (AFD)=60Since AE and EC are angle bisectors of angles A and C so FE is angle bisector of angle ( AFD)Quadrilateral FBEC is cyclic ( F supplement to E) so x= angle (BFE)=30Peter Tran
Problem 698 revisited:With F on AD let EF bisect angle AED.Clearly EA bisects angle BEF andED bisects angle CEFSo triangles AEB and AEF are congruent. So BE = EF.Similarly CE = EF.Hence BE = CE, triangle EBC is isosceles with vertical angle at E = 120 degHence each base angle = x = 30 deg
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The solution is uploaded to the following link:https://docs.google.com/open?id=0B6XXCq92fLJJbG9wSXBldmkweFE