International Mathematical Olympiad

By the conditions we have $BC=DE$, $CT=ET$ and $TB=TD$, so the triangles $TBC$ and $TDE$ are congruent, in particular $\angle BTC=\angle DTE$.

In triangles $TBQ$ and $TES$ we have $\angle TBQ=\angle SET$ and $\angle QTB=180^{\circ }-\angle BTC=180^{\circ }-\angle DTE=\angle ETS$, so these triangles are similar to each other. It follows that $\angle TSE=\angle BQT$ and $$\frac{TD}{TQ}=\frac{TB}{TQ}=\frac{TE}{TS}=\frac{TC}{TS}.$$ By rearranging this relation we get $TD\cdot TS=TC\cdot TQ$, so $C,D,Q$ and $S$ are concyclic. (Alternatively, we can get $\angle CQD=\angle CSD$ from the similar triangles $TCS$ and $TDQ$.) Hence, $\angle DCQ=\angle DSQ$.

Finally, from the angles of triangle $CQP$ we get $$\angle RPQ=\angle RCQ-\angle PQC=\angle DSQ-\angle DSR=\angle RSQ,$$ which proves that $P,Q,R$ and $S$ are concyclic.

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Alternative solution.

As in the previous solution, we note that triangles $TBC$ and $TDE$ are congruent. Denote the intersection point of $DT$ and $BA$ by $V$, and the intersection point of $CT$ and $EA$ by $W$. From triangles $BCQ$ and $DES$ we then have $$\begin{array}{rl}\angle VSW& =\angle DSE=180^{\circ }-\angle SET-\angle TED-\angle EDT\\ & =180^{\circ }-\angle TBA-\angle TCB-\angle CBT=180^{\circ }-\angle QCB-\angle CBQ=\angle BQC=\angle VQW\end{array}$$ meaning that $VSQW$ is cyclic, and in particular $\angle WVQ=\angle WSQ$. Since $$\angle VTB=180^{\circ }-\angle BTC-\angle CTD=180^{\circ }-\angle CTD-\angle DTE=\angle ETW,$$ and $\angle TBV=\angle WET$ by assumption, we have that the triangles $VTB$ and $WTE$ are similar, hence $$\frac{VT}{WT}=\frac{BT}{ET}=\frac{DT}{CT}.$$ Thus $CD\parallel VW$, and angle chasing yields $$\angle RPQ=\angle WVQ=\angle WSQ=\angle RSQ$$ concluding the proof.