Let $ABC$ be an acute triangle. The tangent at $A,B$ of the circumcircle of $ABC$ intersect at $T$. Line $CT$ meets side $AB$ at $D$. Denote by $\Gamma_1,\Gamma_2$ the circumcircle of triangle $CAD$, and the circumcircle of triangle $CBD$, respectively. Let line $TA$ meet $\Gamma_1$ again at $E$ and line $TB$ meet $\Gamma_2$ again at $F$. Line $EF$ intersects sides $AC,BC$ at $P,Q$, respectively. Prove that $EF=PQ+AB$.