Proposed Problem

Click the figure below to see the complete problem 329 about Triangle, Altitudes, Circle, Diameter, Concyclic points.

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Complete Problem 329

Level: High School, SAT Prep, College geometry

## Wednesday, July 29, 2009

### Problem 329. Triangle, Altitudes, Circle, Diameter, Concyclic points

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BC is perpendicular bisector of EF, so BF = BE

ReplyDeleteBy the same reason, BM = BN

Triangle BFC is a right triangle and FD is altitude, so BF^2 = BD*BC

By the same reason, BN^2 = BG*BA.

But BD*BC = BG*BA because angles ADC and AGC are right angles, and four points A, C, D, G are concyclic

Not only are points FMEN concyclic, but B is the center of the circle!

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