Balkan Mathematical Olympiad

Solution: Let the line through $M$ parallel to $AB$ meet the segments $AD,DH,BC,CH$ at points $K,P,L,Q$ respectively. Triangle $HPQ$ is isosceles, so $MP=MQ$. Now from $$\frac{MP}{BH}=\frac{DM}{DB}=\frac{KM}{AB}\text{and}\frac{MQ}{AH}=\frac{CM}{CA}=\frac{ML}{AB}$$ we obtain $\frac{AH}{HB}=\frac{KM}{ML}$. Let the lines $AD$ and $BC$ meet at point $S$ and let the line $SM$ meet $AB$ at $H^{\prime }$. Then $\frac{A{H}^{\prime }}{H^{\prime }B}=\frac{KM}{ML}=\frac{AH}{HB}$, so $H^{\prime }\equiv H$, i.e. $S$ lies on the line $MH$. The quadrilateral $ABCD$ is not a trapezoid, so $AH\ne BH$. Consider the point $A^{\prime }$ on the ray $HB$ such that $H{A}^{\prime }=HA$. Since $\angle S{A}^{\prime }M=\angle SAM=\angle SBM$, quadrilateral $A^{\prime }BSM$ is cyclic and therefore $\angle ABC=\angle A^{\prime }BS=\angle A^{\prime }MH=\angle AMH=90^{\circ }-\angle BAC$, which implies that $\angle ACB=90^{\circ }$.