MathDB
Iranian geometry olympiad 2014(4)

Source: Iranian geometry olympiad 2014

February 25, 2015
geometryIran

Problem Statement

Let ABCABC be an acute triangle.A circle with diameter BCBC meets ABAB and ACAC at EE and FF,respectively. MM is midpoint of BCBC and PP is point of intersection AMAM with EFEF. XX is an arbitary point on arc EFEF and YY is the second intersection of XPXP with a circle with diameter BCBC.Prove that XAY=XYM \measuredangle XAY=\measuredangle XYM . Author:Ali zo'alam , Iran