The triangle midline theorem, also called the midsegment theorem, states that the line segment connecting the midpoints of two sides of a triangle is parallel to and half the length of the third side. This may be trivially proven via triangle similarity (SAS similarity condition) and the corresponding angles postulate. More interesting, however, is to prove it using triangle congruence.

Let *D* and *E* be the midpoints of sides *AB* and *AC*, respectively, of ∆*ABC*. Let us extend segment *DE* past *E* to point *F* such that *DE*=*EF*, and let us draw *CF*.

Since *DE*=*EF*, *AE*=*EC*, and vertical angles ∠*AED* and ∠*CEF* are congruent, we see by the SAS condition that ∆*ADE*≅∆*CFE*. Thus, *CF*=*AD*=*BD*. Also, ∠*FCE*≅∠*DAE*; but since these are alternate interior angles for lines and cut by transversal , we see that . But then the quadilateral *BCFD* has a pair of opposite sides, *BD* and *CF*, which are of equal length and parallel, so it is therefore a parallelogram, and so . And since opposite sides of a parallelogram have equal length, *DF*=*BC*, and so *DE*=½*DF*=½*BC*.

We can also prove a similar theorem: that the line through the midpoint of one side of a triangle parallel to a second side of the triangle bisects the third side, and that the segment of that line inside the triangle is one-half as long as the parallel side. As with the midpoint theorem, it is trivial to prove with triangle similarity (this time the AA condition) and the corresponding angles postulate. So, instead, we use a similar construction as above.

Let *D* be the midpoint of side *AB* of ∆*ABC*, and let *E* be the point where the line through *D* parallel to *BC* intersects *AC*. Construct the line through *C* parallel to *AB*, and let *F* be the point where it intersects

Then *BCFD* is a parallelogram, and since opposite sides of a parallelogram have the same length, *BD*=*CF* and *BC*=*DF*. And so *CF*=*BD*=*AD*. And since ∠*ADE* and ∠*CFE* are alternate interior angles for lines and cut by transversal , they are congruent. Similarly, ∠*DAE*≅∠*CFE*, and so, by the ASA condition, ∆*ADE*≅∆*CFE*. Thus, *AE*=*EC*, and *E* is thus the midpoint of *AC*. We see also that *DE*=*EF*, and since *DF*=*BC*, thus *DE*=½*DF*=½*BC*.

Using both of these theorems together, we can prove a third: that a midline (mid-segment) of a triangle and the triangle median that intersects it bisect each other.

Let *M*_{A}, *M*_{B} and *M*_{C} be the midpoints of sides *BC*, *AC* and *AB*, respectively, of ∆*ABC*. Thus, is a midline of ∆*ABC*, and a median. Let *P* be the point where they intersect.

By the midline theorem, and *M*_{B}M_{C}=½*BC*.

This means, then, that , and so by our second theorem above with regards to ∆*ABM*_{A}, we see that *P* must be the midpoint of , and *M*_{C}P=½*BM*_{A}.

Similarly, our second theorem applied to triangle *AM*_{A}C establishes that *M*_{B}P=½*CM*_{A}. But *BM*_{A}=*CM*_{A}, and so *M*_{B}P=*M*_{C}P, and *P* is the midpoint of as well.