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Belarusian Mathematical Olympiad

Geometry Difficulty 6.1 National Olympiad Prove it Belarus

The bisectors of angles $\angle A$ and $\angle C$ of a convex quadrilateral $ABCD$ meet at the point $E$ , and the bisectors of angles $\angle B$ and $\angle D$ meet at the point $F$ ( $E$ and $F$ lie in the interior of $ABCD$ ). The point $M$ is the midpoint of the segment $EF$ . The points $H_1, H_2, H_3$ and $H_4$ are the foots of the perpendiculars from $M$ to the sides $AB, BC, CD$ and $AD$ respectively.
Prove that $MH_1 + MH_3 = MH_2 + MH_4$ .

Solution

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,
$MH_1 = \frac{x_1 + x_2}{2}, \quad MH_2 = \frac{x_2 + y_1}{2}, \quad MH_3 = \frac{y_1 + y_2}{2} \quad \text{and} \quad MH_4 = \frac{x_1 + y_2}{2}.$
$\text{Therefore, } MH_1 + MH_3 = \frac{x_1 + y_1 + x_2 + y_2}{2} = MH_2 + MH_4.$

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