Triangle midsegment theorem

Dependencies: SAS congruence, Equal alternate angles ⇒ parallel, Parallelogram tests, Parallelogram properties (opposite sides equal, diagonals bisect each other).

Statement

Let $\triangle ABC$ be given, and let $D$, $E$ be the midpoints of $AB$, $AC$ respectively. Then the segment $DE$ joining the two midpoints (called the midsegment of $\triangle ABC$ relative to side $BC$) satisfies

That is, the midsegment is parallel to the third side and exactly half its length.