There is a triangle $ABC$. Its inscribed circle touches the sides $AB$, $BC$, $CD$ in points $N$, $K$ and $M$ respectively. The lines $MN$ and $MK$ intersect the exterior angle $B$ bisector in points $R$ and $S$. How can I prove that the lines $RK$ and $SN$ intersect at the circle?
Since $$\measuredangle SBK=\measuredangle RBN=\measuredangle NMK=90^{\circ}-\frac{\beta}{2},$$ we see that quadrilaterals $RBKM$ and $MNBS$ are cyclics.
Thus, $$180^{\circ}-\measuredangle NIK=\measuredangle SRI+\measuredangle RSI=\measuredangle BMK+\measuredangle BMN=\measuredangle NMK,$$ which says that $MNIK$ is cyclic and we are done!