MathDB

2001.4

Part of Champions Tournament Seniors - geometry

Problems(1)

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

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

5/20/2022
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.
pentagongeometryequal anglesequal segmentsChampions Tournament