Balti Tee 2020 · Ülesanne 12

W.l.o.g. let $AB < AC$. We will prove that triangles $KXS$ and $SYL$ are congruent by SAS, which will finish the proof. As $BX$ and $CY$ are angle bisectors, we obtain:

$$\frac{1}{2} C\hat{A}B = C\hat{X}S = C\hat{X} + \hat{X}S = \frac{1}{2} C\hat{X}A + \hat{X}S.$$

This implies $\hat{X}S = \frac{1}{2} A\hat{Y}B = \hat{Y}B$ and therefore $SX = YB$. Note that $BY = YA$, hence $Y$ is the midpoint of the hypotenuse $BL$ in $\triangle ABL$. Thus $SX = YB = YL$. Similarly, we get $SY = XK$. Finally, as $S$ is the midpoint of arc $C\hat{A}B$, we obtain $\angle SXC = \angle BYS$, thus $\angle KXS = \angle SYL$, finishing the proof of congruency.