Question 181
Let $ABC$ be a triangle, $M$ the midpoint of the segment $BC$ and $D$ a point on the line $AB$ such that $B$ lies between $A$ and $D$. Let $E$ be a point on the other side of the line $CD$ than $B$ such that $\angle EDC = \angle ACB$ and $\angle DCE = \angle BAC$. Let $F$ be the intersection of $CE$ with the parallel to $DE$ through $A$, and let $Z$ be the intersection of $AE$ and $DF$. Show that the lines $AC$, $BF$ and $MZ$ meet in one point.