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

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

By the Midline Theorem:

$BA : BD = 2 : 1$

$BC : BE = 2 : 1$

and:

$AC \parallel DE$

From Parallelism implies Equal Alternate 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.

$\blacksquare$