Question 15
In the parallelogram $ABCD$ point $G$ is chosen on side $AB$. Consider the circle through $A$ and $G$ that is tangent to the extension of $CB$ beyond $B$ at point $P$. The extension of $DG$ beyond $G$ intersects the circle at $L$. If the quadrilateral $GLBC$ is cyclic, prove that $AB = PC$.