Question 123
Let $ABC$ be a triangle. Points $D$ and $E$ lie on the line $AB$ such that $D-A-B-E$, $AD = AC$ and $BE = BC$. The bisectors of the interior angles at the vertices $A$ and $B$ meet the opposite sides at the points $P$ and $Q$, respectively, and the circumcircle of the triangle $ABC$ at the points $M$ and $N$, respectively. The line joining the point $A$ with the centre of the circumcircle of the triangle $BME$ and the line joining the point $B$ with the centre of the circumcircle of the triangle $AND$ meet at a point $X$, $X \neq C$. Prove that $CX \perp PQ$. (Dušan Djukić)