If $C$ is right-angled then $L=B$ and we're done. So assume $C$ is not right-angled. Reflect $C$ over $M$ to get $C'$. Then $C'ACB$ is a parallelogram, so $C'ACL$ is a trapezoid. Let $D,E$ be the projections of $C',A$ onto line $BC$. Then $C'C=2MC=AL$ and $C'D=AE$, so $DC=LE$. If we call $B$ "to the left" of $C$, then $D$ is to the left of $E$ and $L$ is to the left of $C$, so the points are in the order $D,L,C,E$ or $L,D,E,C$. But the second case violates the fact that $C'ACB$ is a parallelogram that is not a rectangle, so the order is $D,L,C,E$. So $DL=CE$, and $C'L=CA$. So our trapezoid is isosceles, and $\angle ALC=\angle C'CL=45^{\circ}\Rightarrow AL\perp CM$