Team Selection Test for IMO 2023

The oriented angle equalities $\angle QXA=\angle KBA=\angle BPA$ and $\angle XQA=\angle PBA$ imply the oriented similarity $AXQ\sim APB$, and the oriented angle equalities $\angle XTA=\angle LBA$ and $\angle TXA=\angle BLA$ imply the oriented similarity $AXT\sim ALB$. If we put a C-coordinate system on the plane such that $A$ is at the origin, these similarities would be expressed as $bx=pq=tl$. Then letting $s:=tb/q=tp/x$, one gets the oriented similarities $SBP\sim TQX$ and $STP\sim BQL$. Consequently one gets $\angle PST=\angle LBQ=\angle LBK+\angle KBQ=\angle KXB+\angle BPQ=\angle PBT$, hence $S$ is on the circle $(BPT)$, hence $\angle XTQ=\angle PSB=\angle PTB$.