Math Core

Connect the midpoints of two sides of any triangle and something surprising happens: the segment you draw is always parallel to the third side and exactly half as long. That one fact, the Triangle Midsegment Theorem, lets you find unknown lengths, measure distances you can't reach directly, and prove facts about triangles using coordinates.

What a midsegment is

Every triangle has three sides, and every side has a midpoint. A segment that joins two of those midpoints is called a midsegment.

Definition

Midsegment of a triangle

A midsegment of a triangle is a segment whose endpoints are the midpoints of two sides of the triangle. Every triangle has exactly three midsegments.

In the triangle below, $D$ is the midpoint of $\overline{AC}$ and $E$ is the midpoint of $\overline{BC}$ , so $\overline{DE}$ is a midsegment. The side it does not touch, $\overline{AB}$ , is the side it is compared to.

D is the midpoint of AC and E is the midpoint of BC. The dashed segment DE is a midsegment.

Look closely: $\overline{DE}$ runs in the same direction as $\overline{AB}$ , and it looks about half as long. That isn't a coincidence.

Triangle Midsegment Theorem

The segment joining the midpoints of two sides of a triangle is

parallel to the third side, and
half as long as the third side.

If $D$ and $E$ are the midpoints of $\overline{AC}$ and $\overline{BC}$ , then $\overline{DE} \parallel \overline{AB}$ and $DE = \dfrac{1}{2}AB$ .

Why it works: a coordinate proof

Coordinates make the theorem easy to prove for every triangle at once. Place the triangle so one vertex is at the origin and one side lies on the $x$ -axis. Using even numbers keeps the midpoints free of fractions:

A(0, 0), \qquad B(2a, 0), \qquad C(2b, 2c).

By the Midpoint Formula, the midpoint of $\overline{AC}$ is $D(b, c)$ and the midpoint of $\overline{BC}$ is $E(a + b, c)$ .

Parallel: $D$ and $E$ have the same $y$ -coordinate, $c$ , so $\overline{DE}$ is horizontal (slope $0$ ). Side $\overline{AB}$ lies on the $x$ -axis, which is also horizontal. So $\overline{DE} \parallel \overline{AB}$ .
Half as long: $DE = (a + b) - b = a$ , while $AB = 2a - 0 = 2a$ . So $DE = \dfrac{1}{2}AB$ .

Because $a$ , $b$ and $c$ can be any numbers, this covers every possible triangle.

Worked example: Using the theorem to find lengths

In $\triangle PQR$ , $S$ is the midpoint of $\overline{PR}$ and $T$ is the midpoint of $\overline{QR}$ .

If $PQ = 17$ , find $ST$ .
If $ST = 7.5$ , find $PQ$ .

S and T are the midpoints of PR and QR.

Solution. $\overline{ST}$ is the midsegment opposite side $\overline{PQ}$ .

The midsegment is half the third side: $ST = \dfrac{1}{2}(17) = 8.5$ .
The third side is twice the midsegment: $PQ = 2(7.5) = 15$ .

Setting up equations

Often the lengths are given as expressions. The relationship is always the same: third side $=$ 2 $\times$ midsegment. Write that equation, solve, and then substitute back to answer the question that was actually asked.

Worked example: A midsegment with algebra

In $\triangle ABC$ , $M$ and $N$ are the midpoints of $\overline{AC}$ and $\overline{BC}$ . If $MN = 2x + 1$ and $AB = 5x - 4$ , find $MN$ and $AB$ .

M and N are the midpoints of AC and BC.

Solution. The third side is twice the midsegment:

\begin{aligned} AB &= 2 \cdot MN \\ 5x - 4 &= 2(2x + 1) \\ 5x - 4 &= 4x + 2 \\ x &= 6 \end{aligned}

So $MN = 2(6) + 1 = 13$ and $AB = 5(6) - 4 = 26$ . Check: $26$ is twice $13$ .

Common mistake

Put the $2$ on the correct side. The third side is the long one, so it equals twice the midsegment. Writing $MN = 2 \cdot AB$ would make the midsegment longer than the side, which is impossible. If your answer has the midsegment longer than the third side, you've flipped the relationship.

Parallel means angle facts too

Since a midsegment is parallel to the third side, every tool from parallel lines applies. In the first figure, $\overline{DE} \parallel \overline{AB}$ with transversal $\overline{AC}$ , so $\angle CDE \cong \angle CAB$ (corresponding angles). Likewise $\angle CED \cong \angle CBA$ . This gives you a quick way to find angle measures inside the small triangle at the top.

Checking on a coordinate grid

On a grid you can verify both parts of the theorem with the Slope Formula and the Distance Formula.

Worked example: Verifying the theorem with coordinates

Triangle $ABC$ has vertices $A(0, 0)$ , $B(6, 6)$ and $C(10, 2)$ . $M$ is the midpoint of $\overline{AB}$ and $N$ is the midpoint of $\overline{AC}$ . Show that $\overline{MN} \parallel \overline{BC}$ and $MN = \dfrac{1}{2}BC$ .

Triangle ABC with M the midpoint of AB and N the midpoint of AC.Open in grapher →

Solution. Midpoints: $M = \left(\dfrac{0 + 6}{2}, \dfrac{0 + 6}{2}\right) = (3, 3)$ and $N = \left(\dfrac{0 + 10}{2}, \dfrac{0 + 2}{2}\right) = (5, 1)$ .

Slopes: $\overline{MN}$ has slope $\dfrac{1 - 3}{5 - 3} = -1$ , and $\overline{BC}$ has slope $\dfrac{2 - 6}{10 - 6} = -1$ . Equal slopes, so the segments are parallel.

Lengths: $MN = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$ and $BC = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}$ . Since $2\sqrt{2}$ is half of $4\sqrt{2}$ , the theorem checks out.

The midsegment triangle

Draw all three midsegments and they form a smaller triangle inside, called the midsegment triangle.

The three midsegments DE, EF and FD form the midsegment triangle, which splits ABC into four congruent triangles.

Each side of $\triangle DEF$ is half of a side of $\triangle ABC$ , so:

The perimeter of the midsegment triangle is half the perimeter of the original.
The four small triangles are congruent (by SSS, since each has sides equal to half of $AB$ , $BC$ and $CA$ ). So each one, including the midsegment triangle, has one-fourth of the original area.

Tip

A quick sanity check: the midsegment triangle's perimeter is simply half the big perimeter. You don't need to know which midsegment matches which side. If a triangle has sides $10$ , $14$ and $18$ (perimeter $42$ ), its midsegment triangle has sides $5$ , $7$ and $9$ (perimeter $21$ ).

Practice

Practice 1

In $\triangle ABC$ , $\overline{DE}$ joins the midpoints of $\overline{AC}$ and $\overline{BC}$ . If $DE = 9$ , what is $AB$ ?