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Perpendicular to bases drawn through P meets line BQ at K

Source: Tuymaada 2009, Junior League, Second Day, Problem 2

July 19, 2009
geometrytrapezoidgeometric transformationhomothetyangle bisectorgeometry unsolved

Problem Statement

M M is the midpoint of base BC BC in a trapezoid ABCD ABCD. A point P P is chosen on the base AD AD. The line PM PM meets the line CD CD at a point Q Q such that C C lies between Q Q and D D. The perpendicular to the bases drawn through P P meets the line BQ BQ at K K. Prove that \angle QBC \equal{} \angle KDA. Proposed by S. Berlov
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