Silk Road Mathematics Competition

Let $BC=a$, $AC=b$, $AB=c$ and $I$ be the incenter of the triangle $ABC$.



From the point $I$ let us draw parallel lines to $AB$ and $BC$, intersecting $AK$ and $KC$ at $P$ and $Q$, respectively. In the triangle $IPQ$ we have $\angle PIQ=\angle ABC$, $IQ=s-c$, $IP=s-a$. Since $\angle NBL=\angle ABC$, $NB=s-a$, $LB=s-c$, it follows that $\angle PIQ=\angle NBL$.

On the ray $IP$ let's choose a point $L_1$ satisfying $I{L}_1=IQ$, and on the ray $IQ$ choose a point $N_1$, such that $I{N}_1=IP$. Then $\Delta I{N}_1{L}_1=\Delta NBL$, $BL||I{L}_1$, $BN||I{N}_1$ and it follows that $L_1{N}_1||LN$.

$IPKQ$ is cyclic. Therefore, $\angle IKP=\angle IQP$, $$\angle KIP+\angle I{L}_1{N}_1=\angle KIP+\angle IQP=\angle KIP+\angle IKP=90^{\circ }$$ and it follows that $IK\perp L_1{N}_1$ and $IK\perp LN$.