69th Belarusian Mathematical Olympiad

Since $E$ lies on the bisector of $\angle BAD$ ($\angle BCD$), the point $E$ is equidistant from the sides $AB$ and $AD$ (respectively, $BC$ and $CD$). A similar statement holds for the point $F$. Denote the distances from the point $E$ to the lines $AB$ and $BC$ by $x_1$ and $y_1$, respectively, and the distances from the point $F$ to the lines $AB$ and $AD$ be $x_2$ and $y_2$, respectively. Since the length of the midline in a trapezium is equal to the half-sum of the base lengths, $$M{H}_1=\frac{x_1+x_2}{2},M{H}_2=\frac{x_2+y_1}{2},M{H}_3=\frac{y_1+y_2}{2}\text{and}M{H}_4=\frac{x_1+y_2}{2}.$$ $$\text{Therefore, }M{H}_1+M{H}_3=\frac{x_1+y_1+x_2+y_2}{2}=M{H}_2+M{H}_4.$$