Thursday, July 2, 2009

Problem 313. Circle, Chord, Tangent, Perpendicular, Geometric Mean

Proposed Problem
Click the figure below to see the complete problem 313.

Problem 313.<br />Circle, Chord, Tangent, Perpendicular, Geometric Mean.
See also:
Complete Problem 313
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:45 AM
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1 comments:

  1. Join C & D to E. By the tangent-chord theorem angle BEC = angle CDE and angle AED = angle ECF. This makes triangles EAD & EFC similar since angle EAD=angle EFC=90. Likewise, triangles CBE & EFD are similar. This gives us: b=(EA/CF)c and a=(BE/DF)c as also EA/CF=DE/CE while BE/DF=CE/DE
    We can therefore say that a*b =(EA/CF)c*(BE/DF)c
    or ab =c^2*(EA/CF)*(BE/DF)=c^2*(DE/CE)*(CE/DE)=c^2. QED

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