Question 32
Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA = FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA = DC$ and $AC$ is the bisector of $\angle DAB$. Point $E$ is chosen so that $EA = ED$ and $AD$ is the bisector of $\angle EAC$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram (with $AM \parallel EX$ and $AE \parallel MX$). Prove that the lines $BD$, $FX$ and $ME$ are concurrent.