Question 37
Let $P$ be a point in the exterior of a circumference $\Gamma$, and let $PA$ be one of the tangents from $P$ to $\Gamma$. The line $l$ passes through $P$ and intersects $\Gamma$ in $B$ and $C$, with $B$ between $P$ and $C$. Let $D$ be the point symmetric to $B$ with respect to $P$. Let $\omega_1$ and $\omega_2$ be the circumferences circumscribed to the triangles $DAC$ and $PAB$ respectively; $\omega_1$ and $\omega_2$ intersect in $E \neq A$. The line $EB$ intersects $\omega_1$ in another point $F$. Prove that $CF = AB$.