Perpendicular Bisector of Triangle is Altitude of Medial Triangle
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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.
$\blacksquare$