Let $ABCDE$ be a convex pentagon such that $BC=DE$. Assume that there is a point $T$ inside $ABCDE$ with $TB=TD$, $TC=TE$, and $\angle ABT=\angle TEA$. Let line $AB$ meet lines $CD$ and $CT$ at points $P$ and $Q$, respectively. Assume that the points $P,B,A,Q$ occur on their line in that order. Let line $AE$ meet lines $CD$ and $DT$ at points $R$ and $S$, respectively. Assume that the points $R,E,A,S$ occur on their line in that order.

Prove that the points $P,S,Q,R$ lie on a circle.

Step 1: The conditions $TB=TD$ and $TC=TE$ mean $T$ lies on the perpendicular bisectors of $BD$ and $CE$. Combined with $BC=DE$, triangles $TBC$ and $TDE$ are congruent (SSS: $TB=TD$, $TC=TE$, $BC=DE$).

Step 2: The congruence gives $\angle TBC=\angle TDE$ and $\angle TCB=\angle TED$. The condition $\angle ABT=\angle TEA$ adds a symmetric angle relation at the 'outer' vertices.

Step 3 (Concyclicity via angles): To show $P,S,Q,R$ are concyclic, we show $\angle QPR+\angle QSR=180^{\circ }$ (or $\angle QPR=\angle QSR$, depending on configuration). Using the angle relations from Steps 1-2 and the collinearity conditions, chase angles at $P$ and $S$ through the triangles formed by the intersections.

Step 4: The key is that the congruence $△TBC\cong △TDE$ creates a rotational symmetry about $T$ that maps line $AB$ to line $AE$ (after accounting for $\angle ABT=\angle TEA$). This symmetry maps $P\leftrightarrow R$ and $Q\leftrightarrow S$ (roughly), establishing the cyclic quadrilateral. $■$