## Thursday, June 4, 2009

### Problem 297: Intersecting Circles, Chord, Secant, Radius, Angle, Perpendicular

Proposed Problem
Click the figure below to see the complete problem 297 about Intersecting Circles, Chord, Secant, Radius, Angle, Perpendicular.

Complete Problem 297
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

1. Join AB and produce CQ to DE and circle Q at a point H and K.

Let 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.

2. Problem 297
Is < 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.

3. 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).

4. Extend AB and ED to meet in F and OQ to intersect Q in G.
<CQG = B-A = <AFD.
Since AB is perpendicular to OQ = QG, hence QC is perpendicular to DE.