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NMTC Problem 2

\(ABCD\) is a quadrilateral inscribed in a circle with centre \(O\). Let \(BD\) bisect \(OC\) perpendicularly. \(P\) is a point on \(AC\) such that \(PC=OC\). \(BP\) cuts \(AD\) at \(E\) and the circle \(ABCD\) at \(F\). Prove that \(PF\) is the geometric mean of \(EF\) and \(BF\).

This a part of my set NMTC 2nd Level (Junior) held in 2014.

Note by Siddharth G
2 years, 9 months ago

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