Sunday, October 17, 2010

Problem 529: Right Trapezoid, Circle, Diameter

Geometry Problem
Click the figure below to see the complete problem 529 about Right Trapezoid, Circle, Diameter.

Problem 529: Right trapezoid.
See also:
Complete Problem 529

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 8:41 PM
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3 comments:

  1. c .t . e. oOctober 18, 2010 at 1:49 AM

    1)draw CG // AB => ▲AGE = ▲FBC => AE = BF
    2)▲AEG ~ ▲AFD ( ang AFD = ang AGE, 90 - FCG & 90 - EGC)
    AD/AF = AG/AE
    AD/AF = BC/AE

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  2. Peter TranOctober 18, 2010 at 11:43 AM

    Let M is the intersection of AD to circle O
    Let N is the projection of point O over chord EF.
    Since NO is the mid base of trapezoid ABCD . N is the midpoint of chord EF and segment AB.
    (1) BN=AN & NE=NF so AE=BF
    (2) Note that CM perpendicular to AD and ABCM is a rectangular so BC= AM
    Consider point A and 2 secants AEF & AMD to circle O
    We have AE.AF=AM.AD so AE.AF=BC.AD

    Peter Tran

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  3. Sumith PeirisJanuary 27, 2019 at 10:20 PM

    If AD cuts the circle at X ABCX is a rectangle and CFEX is therefore an isoceles trapezoid and both results follow easily since Tr.s BCF and AEX are congruent

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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