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

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.

$\blacksquare$