Bulgarian National Mathematical Olympiad

Since $AC\parallel LQ$ and $BC\parallel LP$, we have $S_{ALQ}=S_{CLQ}=S_{PLC}=S_{PLB}$. Let the point $K$ be such that $AKBC$ is a parallelogram. By analogy we have $S_{AKQ}=S_{AKC}=S_{CKB}=S_{PKB}$.

The locus of the points $X$ such that $△AXQ$ and $△PXB$ are oriented in one and the same direction and have equal areas is a straight line $\ell$ passing through the intersecting point of $AQ$ and $BP$. Therefore $\ell \equiv KL$ and $AQ$, $BP$ and $KL$ intersect in a point.

Let $N$ be the midpoint of $CL$. The line $KL$ is the image of $MN$ under homothety of center $C$ and coefficient $2$. Since the point $T$ is symmetric to $C$ with respect to $MN$ we have that $T$ lies on $KL$ which completes the proof.