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Another geometry problem :)

Source: IMO ShortList 1991 Problem 2 (JAP 5)

July 27, 2005
geometrycircumcircletrigonometrycyclic quadrilateralIMO Shortlist

Problem Statement

ABC ABC is an acute-angled triangle. M M is the midpoint of BC BC and P P is the point on AM AM such that MB \equal{} MP. H H is the foot of the perpendicular from P P to BC BC. The lines through H H perpendicular to PB PB, PC PC meet AB,AC AB, AC respectively at Q,R Q, R. Show that BC BC is tangent to the circle through Q,H,R Q, H, R at H H. Original Formulation: For an acute triangle ABC,M ABC, M is the midpoint of the segment BC,P BC, P is a point on the segment AM AM such that PM \equal{} BM, H is the foot of the perpendicular line from P P to BC,Q BC, Q is the point of intersection of segment AB AB and the line passing through H H that is perpendicular to PB, PB, and finally, R R is the point of intersection of the segment AC AC and the line passing through H H that is perpendicular to PC. PC. Show that the circumcircle of QHR QHR is tangent to the side BC BC at point H. H.
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