Sunday, September 12, 2010

Problem 521 Triangle, Squares, Altitude, Rectangles, Areas

Geometry Problem
Click the figure below to see the complete problem 521 about Triangle, Squares, Altitude, Rectangles, Areas.

Problem 521: Triangle, Squares, Altitude, Rectangles, Areas


See also:
Complete Problem 521

Level: High School, SAT Prep, College geometry

2 comments:

  1. We have Angle AHB=angle AMB =90 and quadrilateral ABMH is cyclic
    CMB and CHA are 2 secants from point C to circle ABMH
    So CM.CB=CH.CA
    But CM.CB=S2’ and CH.CA=S2 and S2=S2’
    Similarly we also have S1=S1’ and S3=S3’

    Peter Tran

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  2. ABH = ACP => ▲ABH ~ ▲ACP => AB/AC = AH/AP =>
    AB∙AP = AC∙AH

    ReplyDelete