The angle condition gives us that $\angle ACC_1 = \angle ALB_1 = \angle ABB_1$, which gives us that $BC_1B_1C$ is cyclic. Therefore, Power of the Point at $A$ gives us $AB_1 \cdot AC = AC_1 \cdot AB,$ and so hence $B_1C_1$ is antiparallel to $BC$. However, as it's also parallel, we get that $AB = AC$. Therefore, we also have that $AB_1 = AC_1$, so let's set $AC_1 = AB_1 = x$ and $AB = AC = y$. Then, we have that $AL^2 = AB_1 \cdot AC$ due to the tangent condition, and so hence $AL = \sqrt{xy} \le \frac{x + y}{2} = \frac{AC + AC_1}{2},$ where we used AM-GM. $\square$