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BQ = PE in pentagon <ABC =<AED = 90^o, < BAC< DAE

Source: Champions Tournament (Ukraine) - Турнір чемпіонів - 2001 Seniors p4

May 20, 2022
pentagongeometryequal anglesequal segmentsChampions Tournament

Problem Statement

Given a convex pentagon ABCDEABCDE in which ∠ABC=∠AED=90o\angle ABC = \angle AED = 90^o, ∠BAC=∠DAE\angle BAC= \angle DAE. Let KK be the midpoint of the side CDCD, and PP the intersection point of lines ADAD and BKBK, QQ be the intersection point of lines ACAC and EKEK. Prove that BQ=PEBQ = PE.
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