IMO 2022 Problem 4
The equalities $TB = TD$ and $TC = TE$ place $T$ on the perpendicular bisectors of $BD$ and $CE$, hence $T$
Proposed by: -
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 50s
Problem
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 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.
Exploration
The equalities $TB = TD$ and $TC = TE$ place $T$ on the perpendicular bisectors of $BD$ and $CE$, hence $T$