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.

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
$$\begin{aligned} \angle VSW & = \angle DSE = 180^\circ - \angle SET - \angle TED - \angle EDT \\ & = 180^\circ - \angle TBA - \angle TCB - \angle CBT = 180^\circ - \angle QCB - \angle CBQ = \angle BQC = \angle VQW \end{aligned}$$
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.