Let $\omega$ be the semicircle on diameter $AB$. A line parallel to $AB$ intersects $\omega$ at $C$ and $D$ so that $B$ and $C$ lie on opposite sides of $AD$. The line through $C$ parallel to $AD$ meets $\omega$ again in $E$. Lines $BE$ and $CD$ meet in $F$ and the line through $F$ parallel to $AD$ meets $AB$ in $P$. Prove that $PC$ is tangent to $\omega$.