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KB=KC in triangle ABC, K is intersection of perpendiculars.

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February 13, 2011
geometrygeometric transformationreflectionincenterperpendicular bisectorgeometry unsolved

Problem Statement

In acute triangle ABCABC, an arbitrary point PP is chosen on altitude AHAH. Points EE and FF are the midpoints of sides CACA and ABAB respectively. The perpendiculars from EE to CPCP and from FF to BPBP meet at point KK. Prove that KB=KCKB = KC.
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