Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$. Let $t_{1}$ and $t_{2}$ be the tangents to $\omega_{1}$ and $\omega_{2}$, respectively, at point $A$. Let the second intersection of $\omega_{1}$ and $t_{2}$ be $C$, and let the second intersection of $\omega_{2}$ and $t_{1}$ be $D$. Points $P$ and $E$ lie on the ray $AB$, such that $B$ lies between $A$ and $P$, $P$ lies between $A$ and $E$, and $AE = 2 \cdot AP$. The circumcircle to $\bigtriangleup BCE$ intersects $t_{2}$ again at point $Q$, whereas the circumcircle to $\bigtriangleup BDE$ intersects $t_{1}$ again at point $R$. Prove that points $P$, $Q$, and $R$ are collinear.