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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 871.
Angle MCE = Angle EDF = Angle MED - 90 = Angle MEC So triangle MCE is isosceles and so as triangle BEM for symmetry. Hence BM = EM = CM Q.E.D.
My solution http://www.docstoc.com/docs/153902983/Problem-871
Far more general:If ABCD is cyclic quadrilateral and its diagonals cuts in E, then orthocenter of tr ABE, circumcenter of tr DEC and E itself are colinear.In this particular case, EF contains the orthocenter of tr AED so must contain the circumcenter of tr BEC, but this is M.Done.
Problem 871Let x=<BCA=<BDA=<AEF (perpendicular line) or <MEC=x.Then ME=MC.But <EBC=90-x=<BEM.So BM=EM.Therefore BM=MC.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE