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Iran(Second Round) 2015,second day,problem 4

Source: Iran(Second Round) 2015,second day,problem 4

May 8, 2015
geometry proposedgeometry

Problem Statement

In quadrilateral ABCDABCD , ACAC is bisector of A^\hat{A} and ADC^=ACB^\widehat{ADC}=\widehat{ACB}. XX and YY are feet of perpendicular from AA to BCBC and CDCD,respectively.Prove that orthocenter of triangle AXYAXY is on BDBD.