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King improves upon 8.4

Source: Kyiv City MO 2021 Round 1, Problem 9.5

December 21, 2023
geometry

Problem Statement

Let BMBM be the median of triangle ABCABC in which AB>BCAB > BC. The point PP is chosen so that AB∥PCAB\parallel PC and PM⊥BMPM \perp BM. On the line BPBP, point QQ is chosen so that ∠AQC=90∘\angle AQC = 90^\circ, and points BB and QQ are on opposite sides of the line ACAC. Prove that AB=BQAB = BQ.
Proposed by Mykhailo Shtandenko
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