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Prove that <ABC=3<PBC

Source: Canadian Repêchage 2009: Problem 6

April 25, 2014
geometry proposedgeometry

Problem Statement

Triangle ABCABC is right-angled at CC. AQAQ is drawn parallel to BCBC with QQ and BB on opposite sides of ACAC so that when BQBQ is drawn, intersecting ACAC at PP, we have PQ=2ABPQ = 2AB. Prove that ∠ABC=3∠PBC\angle ABC = 3\angle PBC.
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