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: 20,629 Thanks: 2077 
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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