Triangle midsegment theorem: find missing lengths and perimeter

When a segment joins the midpoints of two sides of a triangle, it tells us something about the third side too. We can use that connection to find missing lengths without measuring the whole triangle. Let’s work through the theorem and two examples.

Learn Geometry with meStep-by-step videos, lesson notes, and practice for area, shapes, and more.

Explore Geometry →

Identify a midsegment of a triangle

A midsegment is a segment joining the midpoints of two sides of a triangle. If $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$, then $AD=DB$ and $AE=EC$. The segment $DE$ is a midsegment.

D is the midpoint of AB, and E is the midpoint of AC. DE is parallel to BC.

Use the triangle midsegment theorem

A triangle’s midsegment is parallel to the third side and half as long as that side. In this triangle, the theorem gives

$\begin{array}{l}DE\parallel BC\\ DE=\frac{1}{2}BC\\ BC=2DE\end{array}$

A triangle has three midsegments, one for each pair of side midpoints. If $F$ is the midpoint of $BC$, then $DE$, $DF$, and $EF$ are all midsegments. Each one is parallel to a different side. Also, $DE=BF=FC$, since each of these lengths is half of $BC$.

Example: Find x from a midsegment of length 8

The midsegment has length $8$, and the parallel side has length $2x+4$. The midsegment theorem gives an equation relating these two lengths.

The purple midsegment is half as long as the parallel side below it.

$\begin{array}{l}8=\frac{1}{2}(2x+4)\\ 8=x+2\\ x=6\end{array}$

The full side has length $2\left(6\right)+4=16$, which is twice the midsegment length $8$.

Example: Find the perimeter using three midsegments

Let $D$, $E$, and $F$ be the midpoints of $AB$, $AC$, and $BC$. Given $DF=4$, $EF=3$, and $FC=5$, find the perimeter of triangle $ABC$.

D, E, and F are midpoints. DF = 4, EF = 3, and FC = 5.

The segment $EF$ is parallel to $AB$, so $AB$ is twice $EF$. The segment $DF$ is parallel to $AC$, so $AC$ is twice $DF$. Since $F$ is the midpoint of $BC$, double $FC$ to get $BC$.

$\begin{array}{l}AB=2\left(3\right)=6\\ AC=2\left(4\right)=8\\ BC=2\left(5\right)=10\\ P=6+8+10=24\end{array}$

Full Geometry course

Learn Geometry with me

We’ll work through area, triangles, circles, and proofs, one step at a time.

Explore my Geometry course →

• Video lessons
• Lesson notes
• Practice problems

Included with membership, along with every other Krista King Math course.