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 segment joining 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 $N$ the midpoint of $AC$, then:

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

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:

We'll prove both parts using SAS similarity.

Step 3: SAS Similarity

Consider triangles $AMN$ and $ABC$:

• $AM/AB=1/2$ ($M$ is the midpoint of $AB$)
• $AN/AC=1/2$ ($N$ is the midpoint of $AC$)
• $\angle A$ is shared

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

Step 4: $MN\parallel BC$

Since $△AMN\sim △ABC$, 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:

The midsegment is half the length of the third side.

The midsegment connecting midpoints of two sides is:

• Parallel to the third side
• Half the length of the third side

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