The Midsegment Theorem

A proof that the segment connecting midpoints of two sides of a triangle is parallel to the third side and half its length.

Keywords: midsegment, midpoints, parallel, similar triangles, SAS similarity

Prerequisites: triangle-similarity · Difficulty: beginner

The midsegment theorem states: the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length.

If $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$, then:

MN \parallel BC \quad \text{and} \quad MN = \frac{1}{2} BC

Proof Strategy

By SAS similarity: $AM/AB = AN/AC = 1/2$ and $\angle A$ is shared. So $\triangle AMN \sim \triangle ABC$ with ratio $1:2$. The corresponding sides give $MN = BC/2$, and equal corresponding angles give $MN \parallel BC$.

The Proof

Step 1: Midpoints and the Midsegment

Start with triangle $ABC$. Mark the midpoints $M$ of side $AB$ and $N$ of side $AC$. Connect them to form the midsegment $MN$.

Step 2: The Midsegment Theorem

The Midsegment Theorem says that the segment $MN$ joining the midpoints of two sides is parallel to the third side and half its length:

MN \parallel BC, \quad MN = \tfrac{1}{2} BC

We'll prove both parts using SAS similarity.

Step 3: SAS Similarity

Consider triangles $AMN$ and $ABC$:

By SAS similarity, $\triangle AMN \sim \triangle ABC$ with ratio $1:2$.

Step 4: $MN \parallel BC$

Since $\triangle AMN \sim \triangle ABC$, corresponding angles are equal: $\angle AMN = \angle ABC$. These are corresponding angles with respect to transversal $AB$, so $MN \parallel BC$.

Step 5: $MN = ½BC$

The similarity ratio is $1:2$, so corresponding sides are in ratio $1:2$. In particular:

\frac{MN}{BC} = \frac{1}{2} \implies MN = \frac{1}{2}BC

The midsegment is half the length of the third side.

Conclusion

The midsegment connecting midpoints of two sides is:

MN \parallel BC, \quad MN = \frac{1}{2}BC

This theorem is used repeatedly in proofs about medians, centroids, and the nine-point circle.