Let $H$ be the orthocenter of triangle $ABC$ and $X\neq A$ be the intersection of line $AA_1$ with $ABC$ circumcircle. Quadrilateral $BCB_1C_1$ is cyclic so $$\measuredangle KXC=\measuredangle ABC=\measuredangle KB_1C\implies\ K, X, B_1, C\text{ are concyclic}\implies AX\cdot AK=AB_1\cdot AC.$$Quadrilateral $B_1CA_1H$ is also cyclic thus $$AH\cdot AA_1=AB_1\cdot AC.$$It's known that $X$ is the reflection of $H$ over $A_1$: $AH=AX-2A_1X$. $$AH\cdot AA_1=AX\cdot AK\iff AA_1\cdot A_1X=AX\cdot\frac{AA_1-AK}{2}\iff AA_1\cdot A_1X=AX\cdot\frac{A_1K}{2}\iff AA_1^2=AX\cdot \left(AA_1-\frac{A_1K}{2}\right).$$Denote by $M$ the midpoint of $A_1K$. We get$$\frac{AA_1}{AX}=\frac{AM}{AA_1}.$$$$ML\perp A_1A\perp BC\implies PL\parallel A_1B\implies \frac{AL}{AB}=\frac{AM}{AA_1}=\frac{AA_1}{AX}\implies \triangle ALA_1\sim\triangle ABX\ (s,a,s).$$Analogously we prove $$ \triangle AMA_1\sim\triangle ACX.$$Therefore $\text{quadrilateral}\ AMA_1L\sim \text{quadrilateral}\ ACXB.$ Now everything's obvious since $\text{quadrilateral}\ ACXB$ is cyclic. QED