Question 43
In a cyclic hexagon $ABCDEF$, $AB\perp BD$ and $|BC|=|EF|$. Let $P=BC\cap AD$ and $Q=EF\cap AD$. Assume that both $P$ and $Q$ lie beyond $D$ on the side opposite $A$. Let $S$ be the midpoint of $AD$, and let $K,L$ be the incenters of triangles $BPS,EQS$, respectively. Prove that $\angle KDL=90^\circ$.
