FINAL ROUND

Let $X$ be the intersection point of the lines $L{C}_1$ and $K{B}_1$. To prove the problem statement it suffices to show that points $X$, $I$ and $A_1$ lie on the same line. Let $\angle CAB=\alpha$, $\angle ABC=\beta$, $\angle BCA=\gamma$. Since $L{C}_1{B}_{1}C$ is an inscribed quadrilateral, we have $$\angle C_{1}LC=180^{\circ }-\angle C_1{B}_{1}C=\angle C_1{B}_{1}A=\angle B_1{C}_{1}A=90^{\circ }-\frac{\alpha }{2}=(\beta +\gamma )/2.$$ Similarly, $\angle B_{1}KB=(\beta +\gamma )/2$. Therefore the triangle $LXK$ is isosceles ($XL=XK$) and $$\angle C_{1}X{B}_1=\angle LXK=180^{\circ }-(\beta +\gamma )=\alpha .$$ Since $I{C}_1\perp AB$ and $I{B}_1\perp AC$, we have $\angle C_{1}I{B}_1=180^{\circ }-\alpha$.



Therefore $I$, $C_1$, $A$, $X$, $B_1$ lie on the same circle $\omega$. The bisector of the angle $C_{1}A{B}_1$ meets $\omega$ at points $A$ and $I$. So $$\angle C_{1}XI=\angle C_{1}AI=\angle IA{B}_1=\angle IX{B}_1,$$ which yields that $XI$ is the bisector of the angle $C_{1}X{B}_1$. Since the triangle $LXK$ is an isosceles triangle, we see that $XI$ is the altitude of this triangle. On the other hand, $I{A}_1\perp BC$, so points $X$, $I$, $A$ lie on the same line, as required.