Mathematical Olympiad Rioplatense

Let $E$ be the second common point of $DA$ and the circle. One can show that $A$ is between $D$ and $E$. Denote $\angle ELG=\alpha$, $\angle GLC=\beta$. Since $GLBC$ is a cyclic quadrilateral by hypothesis, we have $\angle GBC=\angle GLC=\beta$; since $ELGA$ is also cyclic, $\angle DAG=\alpha$. Thus $\alpha$ and $\beta$ are the measures of adjacent angles in a parallelogram, hence $\alpha +\beta =180^{\circ }$. It follows that $E$, $L$ and $C$ are collinear.



Let $\angle LEA=\theta$, then $\angle LGB=\theta$ as $ELGA$ is cyclic. Hence $\angle LGB=\angle LDC=\theta$ due to $AB\parallel CD$. Triangles $EDC$ and $DLC$ are similar as they share an angle at $C$ and $\angle CED=\angle CDL=\theta$. The similitude gives $\frac{CD}{CL}=\frac{CE}{CD}$, or $C{D}^2=CL\cdot CE$. On the other hand, by power of a point, $C{P}^2=CL\cdot CE$. Hence $CD=CP$. Since $CD=AB$ (opposite sides of a parallelogram), we obtain $AB=CP$, as desired.