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China South East Mathematical Olympiad 2013 problem 2

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August 10, 2013
geometrygeometry unsolved

Problem Statement

△ABC\triangle ABC, AB>ACAB>AC. the incircle II of △ABC\triangle ABC meet BCBC at point DD, ADAD meet II again at EE. EPEP is a tangent of II, and EPEP meet the extension line of BCBC at PP. CF∥PECF\parallel PE, CF∩AD=FCF\cap AD=F. the line BFBF meet II at M,NM,N, point MM is on the line segment BFBF, the line segment PMPM meet II again at QQ. Show that ∠ENP=∠ENQ\angle ENP=\angle ENQ
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