Problem

Source: Bosnia and Herzegovina JBMO TST 2018

Tags: geometry



Let $\Gamma$ be circumscribed circle of triangle $ABC $ $(AB \neq AC)$. Let $O$ be circumcenter of the triangle $ABC$. Let $M$ be a point where angle bisector of angle $BAC$ intersects $\Gamma$. Let $D$ $(D \neq M)$ be a point where circumscribed circle of the triangle $BOM$ intersects line segment $AM$ and let $E$ $(E \neq M)$ be a point where circumscribed circle of triangle $COM$ intersects line segment $AM$. Prove that $BD+CE=AM$.