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 \(AB\) intersect 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 \(AE\) intersect \(CD\) 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.
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
\[\begin{aligned} \angle VSW & = \angle DSE = 180^{\circ} - \angle SED - \angle EDS = 180^{\circ}-\angle AET - \angle TED - \angle EDT \\ & =180^{\circ}-\angle TBA-\angle TCB-\angle CBT = 180^{\circ}-\angle QCB-\angle CBQ = \angle BQC = \angle VQW \end{aligned}\]
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 \(\angle T B V=\angle W E T\) by assumption, we have that the triangles \(VTB\) and \(WTE\) are similar, hence\[\frac{VT}{WT} = \frac{BT}{ET} = \frac{DT}{CT}\]
Thus \(CD | VW\), and angle chasing yields\[\angle R P Q=\angle W V Q=\angle W S Q=\angle R S Q\]
concluding the proof.