Triangle is Medial Triangle of Larger Triangle

Theorem

Let $\triangle ABC$ be a triangle.

$\triangle ABC$ is the medial triangle of a larger triangle.

Construction

Consider $\triangle ABC$.

Let $D, E, F$ be constructed so that:

$DE \parallel AC$

$EF \parallel AB$

$DF \parallel BC$

Then $\triangle ABC$ is the medial triangle of $\triangle DEF$.

Proof

By Euclid's Fifth Postulate, it is possible to construct exactly one straight line parallel to each of $AC$, $BC$ and $AC$.

From Parallelism implies Equal Corresponding Angles:

$\angle ABC = \angle ECB = \angle DAB$

$\angle ACB = \angle CBE = \angle CAF$

$\angle CAB = \angle ABD = \angle ACF$

By Triangle Angle-Side-Angle Congruence:

$\triangle ABC = \triangle ABD$

$\triangle ABC = \triangle CFA$

$\triangle ABC = \triangle ECB$

Thus:

$FC = CE$

$DB = BE$

$FA = AD$

The result follows by definition of medial triangle.

$\blacksquare$

Sources

• For a video presentation of the contents of this page, visit the Khan Academy.