Suppose a convex pentagon ABCDE such that BC=DE.If there exists a point T inside ABCDE such

that TB=TD TC=TE and ∠ABT=∠TEA. AB meet CD and CT at point P and Q respectively, with

P,B,A,Q in this order on the same line. AE meet CD and DT at point R and S respectively, with R,E,A,S in this order on the same line.Prove that P,S,Q,R are on the same circle.

译文：

设凸五边形ABCDE满足BC=DE.若在ABCDE内存在一点T使得TB=TD,TC=TE且∠ABT= ∠TEA.直线AB分别与直线CD和CT交于点P和Q，且P,B,A,Q在同一直线上按此顺序排列;直线AE分别与直线CD和DT交于点R和S，且R,E,A,S在同一直线上按此顺序排列.证明：P,S,Q,R 四点共圆.