Perpendicular Bisector of Triangle is Altitude of Medial Triangle

Theorem
Let $\triangle ABC$ be a triangle.

Let $\triangle DEF$ be the medial triangle of $\triangle ABC$.

Let a perpendicular bisector be constructed on $AC$ at $F$ to intersect $DE$ at $P$.

Then $FP$ is an altitude of $\triangle DEF$.

Proof

 * PerpendicularBisectorAltitudeOfMedial.png

Consider the triangles $\triangle ABC$ and $\triangle DBE$.

By construction:
 * $BA : BD = 2 : 1$
 * $BC : BE = 2 : 1$

By Parallel Line in Triangle Cuts Sides Proportionally:
 * $AC \parallel DE$

From Parallelism implies Equal Alternate Interior Angles:
 * $\angle AFP = \angle FPE$

By construction, $\angle AFP$ is a right angle.

Thus $\angle FPE$ is also a right angle.

That is, $FP$ is perpendicular to $DE$.

By construction, $FP$ passes through the vertex $F$ of $\triangle DEF$.

The result follows by definition of altitude of triangle.