Let $X$ be the intersection point of the lines $LC_1$ and $KB_1$. 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 $LC_1B_1C$ is an inscribed quadrilateral, we have
$$\angle C_1LC = 180^\circ - \angle C_1B_1C = \angle C_1B_1A = \angle B_1C_1A = 90^\circ - \frac{\alpha}{2} = (\beta + \gamma)/2.$$
Similarly, $\angle B_1KB = (\beta + \gamma)/2$. Therefore the triangle $LXK$ is isosceles ($XL = XK$) and
$$\angle C_1XB_1 = \angle LXK = 180^\circ - (\beta + \gamma) = \alpha.$$
Since $IC_1 \perp AB$ and $IB_1 \perp AC$, we have $\angle C_1IB_1 = 180^\circ - \alpha$.

Therefore $I$, $C_1$, $A$, $X$, $B_1$ lie on the same circle $\omega$. The bisector of the angle $C_1AB_1$ meets $\omega$ at points $A$ and $I$. So
$$\angle C_1XI = \angle C_1AI = \angle IAB_1 = \angle IXB_1,$$
which yields that $XI$ is the bisector of the angle $C_1XB_1$. Since the triangle $LXK$ is an isosceles triangle, we see that $XI$ is the altitude of this triangle. On the other hand, $IA_1 \perp BC$, so points $X$, $I$, $A$ lie on the same line, as required.