Given a quadrilateral $ABCD$ inscribed in circle $\Gamma$.From a point P outside $\Gamma$, draw tangents $PA$ and $PB$ with $A$ and $B$ as touspoints. The line $PC$ intersects $\Gamma$ at point $D$. Draw a line through $B$ parallel to $PA$, this line intersects $AC$ and $AD$ at points $E$ and $F$ respectively. Prove that $BE = BF$.