MathDB
question 6

Source: iran tst 2014 third exam

May 21, 2014
geometrycircumcircletrigonometryratiosymmetrygeometry proposed

Problem Statement

The incircle of a non-isosceles triangle ABCABC with the center II touches the sides BCBC at DD. let XX is a point on arc BCBC from circumcircle of triangle ABCABC such that if E,FE,F are feet of perpendicular from XX on BI,CIBI,CI and MM is midpoint of EFEF we have MB=MCMB=MC. prove that BAD^=CAX^\widehat{BAD}=\widehat{CAX}