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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, 2 months ago