Question 128
Let $ABCD$ be a parallelogram with $|AC| = |BC|$. A point $P$ is chosen on the ray $AB$ such that $B$ lies between $A$ and $P$. Apart from the point $D$, the circumcircle of triangle $ACD$ and the segment $PD$ also have the point $Q$ in common. Apart from the point $P$, the circumcircle of triangle $APQ$ and the segment $PC$ also have the point $R$ in common.
Prove that the three lines $CD$, $AQ$ and $BR$ meet at one point.
Prove that the three lines $CD$, $AQ$ and $BR$ meet at one point.