Ghd wrote: Let $ABC$ be a triangle with $\angle C=90^{\circ}$, and $K, L $ be the midpoints of the minor arcs $AC$ and $BC$ of its circumcircle. Segment $KL$ meets $AC$,at point $N$. Find angle $NIC$ where $I$is the incenter of $ABC$. We see that since $AL,BK$ bisect $\angle A, \angle B$ respectively and $AB$ sub tends an angle of $90$ on both, $KL$ is infact the intouch chord by the right angles on the intouch chord Lemma. Thus, our answer is $45$.