Proposed Problem
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Complete Problem 313
Level: High School, SAT Prep, College geometry
Thursday, July 2, 2009
Problem 313. Circle, Chord, Tangent, Perpendicular, Geometric Mean
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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
ReplyDeleteWe 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
Connect EC and ED.
ReplyDeleteWe see that triangle EBC and DFE are similar therefore :
a/EC=c/DE
(1.) a/c=EC/DE
and triangle EAD and CFE are similar therefore
c/EC=b/DE
(2.) c/b=EC/DE
From equations 1 and 2 we have
c/b=a/c
c^2=ab