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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 799.
http://img19.imageshack.us/img19/9383/problem799.pngDraw angles and lines per attached sketchWe have ∆ABE and ∆ECF are isosceles trianglesQE=QC (Q is circumcenter of ∆CEF)AB=CD=BE∠EQH=∠CQH=∠EFC= α∠ECD=2 α ( external angle of ∆ ECF)∠QCD= α +90 and ∠QEB=external angle of EQH= α +90So ∆BQE congruence to ∆DQC ( Case SAS) =>∠ BQD=2 α = ∠BCDSo quadrilateral BDCQ is cyclic
Our target is to prove that B,Q,C,D concyclic.Let x be the angle FAD (=angle BAE = angle BEA = angle FEC = EFC) [from previous problem797]Since EFC is isos., QC is the perpendicular bisector of EF and angle QCE = 90-x. Then angle QCD = 90 + x. Now observe that QC = QE BE = BA = CD Angle QCD = 90 + x = 180 - (90+x) = 180 - (angle QEC) = angle QEBHence triangle QEB is congruent to triangle QCD.Therefore angle QBC = angle QBE = angle QDC.By the converse of angle in the same segment, it is proved.
∆'s ABE, CEF are isosceles and CQ ⊥ bisector of EFimply ∠QEB = Π - ∠QEC = Π - ∠QFC = Π - ∠QCF = ∠QCDSo ∆QBE ≡ ∆QDC, ∠QBC = ∠QBE = ∠QDCHence Q, B, D, C are concyclic
Also BD is tangential to circle BEQ