Wednesday, December 19, 2012

Problem 833: Parallelogram, Diagonal, Similarity, Triangle, Proportion

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 833.

Online Geometry Problem 833: Parallelogram, Diagonal, Similarity, Proportion

Posted by Antonio Gutierrez at 7:20 AM
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2 comments:

  1. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 19, 2012 at 7:57 AM

    Construim paralela prin punctul C la dreapta BE si notam cu M intersectia acestei paralele cu dreapta AD.Va rezulta ca patrulaterul BMNC este paralelogram =>BC=EM=AD.Aplicand teorema lui Thales in triunghiul ACM cu EF paralela cu CM va rezulta AF/FC=AE/EM si in final AF/FC=AE/AD

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  2. Peter TranDecember 19, 2012 at 12:29 PM

    Note that triangles AFE and CFB are similar ( Case AA)
    so AE/BC=AF/FC
    but BC=AD so AE/AD=AF/FC

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