Iranian Mathematical Olympiad

Suppose that $L$ and $K$ lie on the circumcircle of $ABD$ and $ADC$, respectively.

Note that $\angle LAB=90^{\circ }-\angle ABL=90^{\circ }-\angle ADL=90^{\circ }-\angle ACB$. Similarly we have $\angle CAK=90^{\circ }-\angle CBA$. Summing these two implies that $\angle LAK=2\angle BAC$. Denote by $T$ the intersection of $BL$, $CK$, then $$\angle TBC=180^{\circ }-\angle CBA-\angle ABL=180^{\circ }-\angle CBA-\angle ACB=\angle BAC,$$ and similarly $\angle TCB=\angle BAC$. Therefore $ABC$ is an isosceles triangle and $\angle CTB=180^{\circ }-2\angle BAC$. Similarly, one can show that $\angle CSB=180^{\circ }-2\angle BAC$ and so quadrilateral $BCTS$ is cyclic. Properties of $S$ imply that $A$ is the $S$-excenter of the triangle $BCS$ and so $SA$ is the angle bisector of $\angle BSC$. Let $M$ be the midpoint of the arc $BC$ (the one that does not contain $S$) in the circumcircle of $BCS$. Clearly $M$ lies on $SA$. On the other hand, since triangle $BTC$ is isosceles, $MT$ is a diagonal of the circumcircle of $BCS$. Thus, we can conclude that $$\angle AST=\angle MST=90^{\circ }=\angle AKT=\angle ALT.$$ This implies $AKLS$ is cyclic as desired. ■