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, it is possible to construct exactly one straight line parallel to each of $AC$, $BC$ and $AC$.


 * TriangleIsMedial.png

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 Equality:
 * $\triangle ABC = \triangle ABD$
 * $\triangle ABC = \triangle CFA$
 * $\triangle ABC = \triangle ECB$

Thus:
 * $FC = CE$
 * $DB = BE$
 * $FA = AD$

The result follows by definition of medial triangle.