Proposed Problem
Click the figure below to see the complete problem 297 about Intersecting Circles, Chord, Secant, Radius, Angle, Perpendicular.
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Complete Problem 297
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Thursday, June 4, 2009
Problem 297: Intersecting Circles, Chord, Secant, Radius, Angle, Perpendicular
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Join AB and produce CQ to DE and circle Q at a point H and K.
ReplyDeleteLet angle BAC=x, then angle EDC=x and angle BKH=x. Since KC is the diameter of circle Q, angle KBC=90. Also triangle CKB is similar to triangle CDH, so angle DHC=90, CQ is perpendicular to DE.
Problem 297
ReplyDeleteIs < AEB=<AQB=2.<ACB=2.ECB.But < AEB=<ECB+<EBC, then <EBC=<ECB so BE=EC.Then EQ is perpendicular to BC. Similar DQ is perpendicular to EC.Then point Q
Is orthocenter triangle DEC.Therefore QC is perpendicular to DC.
If Q is not on O, the result is still valid (as shown by the proof of anonymous in 2009 which does not use the fact that Q is on O).
ReplyDeleteExtend AB and ED to meet in F and OQ to intersect Q in G.
ReplyDelete<CQG = B-A = <AFD.
Since AB is perpendicular to OQ = QG, hence QC is perpendicular to DE.