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Source: Iran Second Round 2015 - Problem 3 Day 1

May 7, 2015
geometry2015

Problem Statement

Consider a triangle ABCABC . The points D,ED,E are on sides AB,ACAB,AC such that BDECBDEC is a cyclic quadrilateral. Let PP be the intersection of BEBE and CDCD. HH is a point on ACAC such that ∠PHA=90∘\angle PHA = 90^{\circ}. Let M,NM,N be the midpoints of AP,BCAP,BC. Prove that: ACD∼MNH ACD \sim MNH .
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