Let the line through $M$ parallel to $A B$ meet the segments $A D, D H, B C, C H$ at points $K, P, L, Q$, respectively. Triangle $H P Q$ is isosceles, so $M P = M Q$. Now from

$$\frac{M P}{B H} = \frac{D M}{D B} = \frac{K M}{A B} \quad \text{and} \quad \frac{M Q}{A H} = \frac{C M}{C A} = \frac{M L}{A B}$$

we obtain $\frac{A H}{H B} = \frac{K M}{M L}$.
Let the lines $A D$ and $B C$ meet at point $S$ and let the line $S M$ meet $A B$ at $H'$. Then $\frac{A H'}{H' B} = \frac{K M}{M L} = \frac{A H}{H B}$, so $H' \equiv H$, i.e., $S$ lies on the line $M H$.
The quadrilateral $A B C D$ is not a trapezoid, so $A H \neq B H$. Consider the point $A'$ on the ray $H B$ such that $H A' = H A$. Since $\angle S A' M = \angle S A M = \angle S B M$, quadrilateral $A' B S M$ is cyclic and therefore $\angle A B C = \angle A' B S = \angle A' M H = \angle A M H = 90^\circ - \angle B A C$, which implies that $\angle A C B = 90^\circ$.
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