Wednesday, July 29, 2009

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

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

 Problem 329. Triangle, Altitudes, Circle, Diameter, Concyclic points.
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Complete Problem 329
Level: High School, SAT Prep, College geometry

2 comments:

  1. BC is perpendicular bisector of EF, so BF = BE
    By 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

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

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