Problem
Suppose that any straight line passes through the vertex $A$ of a triangle $ABC$. The feet $D$ and $E$ of the perpendicular lines drawn from $B$ and $C$ to the above line are equidistant from the midpoint $M$ of the side $BC$.
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Solution
Let $F$ be the foot of the perpendicular line drawn from $M$ to $DE$. Then,
$$BD∥MF∥CE.$$
Since $M$ is the midpoint of $BC$,
$$DF=FE.$$
Therefore, $M$ lies on the perpendicular bisector of $DE$ and is the vertex of the isosceles triangle $MDE$.
$$∴ \ MD=ME.$$
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Reference
Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.17