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April 23rd, 2010, 02:36 PM   #1
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Circle Chord

Points D, E, and F lie on a circle with center at O. EH is tangent to the circle at E. Chords EF and DJ intersect at K such that DJ is parallel to EH. If DE = 180 and EK = 108, find KF.

It seems that in order to find the solution you need to use Power of a Point and notice that OH is perpendicular to DJ, but I don't know how to get started with using the value of DE.
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April 25th, 2010, 10:51 AM   #2
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Re: Circle Chord

Hi Recipe,
I cannot see how OH is perpendicular to DJ - given that DJ is parallel to EH unless H is at the same point as E

Not sure if we have enough info to go on from your description - can you upload a diagram?
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April 25th, 2010, 07:23 PM   #3
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FE/DE = DE/KE (since triangles DEF, KED are similar), so FE = DE/KE = 180/108 = 300.
Hence KF = FE - KE = 300 - 108 = 192.
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April 26th, 2010, 10:34 AM   #4
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Re: Circle Chord

Agree with solution of skipjack with similarity of triangles but cannot see how OH is perpendicular to DJ as originally posed in question?
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