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.

 Problem 297: Intersecting Circles, Chord, Secant, Radius, Angle, Perpendicular.
See also:
Complete Problem 297
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

2 comments:

  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.

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  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.

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