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<PAQ = 90^o -/2 <BAC if AP = PK, AQ = QK, 3 angle bisectors

Source: TOT 469 1995 Spring J A3 - Tournament of Towns

July 9, 2024
geometryanglesangle bisector

Problem Statement

Let AKAK, BLBL and CMCM be the angle bisectors of a triangle ABCABC, with KK on BCBC. Let PP and QQ be the points on the lines BLBL and CMCM respectively such that AP=PKAP = PK and AQ=QKAQ = QK. Prove that ∠PAQ=90o−12∠BAC.\angle PAQ = 90^o -\frac12 \angle B AC.
(I Sharygin)
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