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Circle ChordPoints 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. |

Re: Circle ChordHi 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? |

FE/DE = DE/KE (since triangles DEF, KED are similar), so FE = DE²/KE = 180²/108 = 300. Hence KF = FE - KE = 300 - 108 = 192. |

Re: Circle ChordAgree 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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