Let $X \ne A$ and $Y \ne C$ be the points where $AR$ and $CS$ intersect $T_3$ respectively. Then $CY \perp QP$ and $AX \perp QP$, so $AX||CY$ and since $AC$ is the diameter of $T_3$ we have $YC=XA$ (symmetry), and therefore $QR=PS$ (symmetry again) P.S. Either I am wrong or this problem is soooooo much obvious

Let $O$ be the center of $T_3$ and $M$ the midpoint of $SR.$ The quadrilateral $ASCR$ is a trapezoid, $O, M$ are the midpoints of it's diagonals, so $OM||AR$ which means $OM\bot RS.$ Hence, $M$ is also the midpoint of $PQ$ and finally $\boxed{PR=QS}$

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Observe that $AR\perp PQ$ and $CS\perp PQ$. Use complex numbers with $T_3$ as the unit circle, then $c=-a$, so$$r=\frac{p+q+a-pq\overline a}{2}$$and$$s=\frac{p+q-a+pq\overline a}{2}.$$We get$$p-r=\frac{p-q-a+pq\overline a}{2}=s-q,$$which implies the problem.