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 April 23rd, 2010, 02:36 PM #1 Member   Joined: Dec 2009 Posts: 66 Thanks: 0 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.
 April 25th, 2010, 10:51 AM #2 Newbie   Joined: Apr 2010 Posts: 14 Thanks: 0 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?
 April 25th, 2010, 07:23 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,973 Thanks: 2224 FE/DE = DE/KE (since triangles DEF, KED are similar), so FE = DE²/KE = 180²/108 = 300. Hence KF = FE - KE = 300 - 108 = 192.
 April 26th, 2010, 10:34 AM #4 Newbie   Joined: Apr 2010 Posts: 14 Thanks: 0 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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