Midline Theorem

Theorem
The midline of a triangle is parallel to the third side of that triangle and half its length.

Proof


Let $\triangle ABC$ be a triangle.

Let $DE$ be the midline of $\triangle ABC$ through $AB$ and $AC$.

Extend $DE$ to $DF$ so $DE = EF$.

As $E$ is the midpoint of $AC$, the diagonals of the quadrilateral $ADCF$ bisect each other.

From Quadrilateral with Bisecting Diagonals is Parallelogram, $ADCF$ is a parallelogram.

By definition of a parallelogram, $AB \parallel CF$.

From Opposite Sides and Angles of Parallelogram are Equal, $AD = CF$.

But $AD = DB$ as $D$ is the midpoint of $AB$.

So $DB = CF$ and $DB \parallel CF$.

From Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel, $BCFD$ is a parallelogram.

Thus also by Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel $DF = BC$ and $DF \parallel BC$.

As $DE = EF$, $DE$ is the midpoint of $DF$ and so $DE = \dfrac 1 2 DF$.

Thus $DE = \dfrac 1 2 BC$ and $DE \parallel BC$.

Hence the result.