parmenides51 wrote: Given triangle $ABC$. Point $M$ is the projection of vertex $B$ to bisector of angle $C$. $K$ is the touching point of the incircle with side $BC$. Find angle $\angle MKB$ if $\angle BAC = \alpha$ (V.Protasov) By Sharigin problem 255 we immediately get the answer $\frac{\pi-\alpha}{2}$: if points $L$ and $N$ are the touching points of the incircle and the sides $AB$ and $AC$, then points $N$, $L$ and $M$ are colinear. Hence $\angle MKB=\angle MLB=\angle ALN=\frac{\pi-\alpha}{2}$.