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<BPD = 2 <CPD wanted, <ВАС =< ВРВ, isosceles AB=AC, BD:DC = 2: 1

Source: Champions Tournament (Ukraine) - Турнір чемпіонів - 2006 Seniors p3

September 6, 2020
geometryisoscelesequal anglesChampions Tournament

Problem Statement

Let ABCABC be an isosceles triangle with AB=ACAB = AC. Let DD be a point on the base BCBC such that BD:DC=2:1BD:DC = 2: 1. Note on the segment ADAD a point PP such that ∠BAC=∠BPD\angle BAC= \angle BPD . Prove that ∠BPD=2∠CPD\angle BPD = 2 \angle CPD.
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