A midsegment of a triangle joins the midpoints of two sides. Every triangle has three. With the midpoint formula from Distance and Midpoint Formulas and the parallel-line facts behind Triangle Sum and Exterior Angle Proofs, we can show that a midsegment always does two surprising things.
Why it works
Put the triangle on a grid in a smart spot: $A(0, 0)$, $B(2a, 0)$ and $C(2b, 2c)$. (Using $2a$, $2b$, $2c$ keeps the midpoints free of fractions.) The midpoint of $\overline{AC}$ is $D(b, c)$ and the midpoint of $\overline{BC}$ is $E(a + b, c)$. $\overline{DE}$ is horizontal, just like $\overline{AB}$, so they are parallel. $DE = (a + b) - b = a$, while $AB = 2a$. Since the triangle was any triangle, this holds for all of them.
Key idea
Triangle Midsegment Theorem: a midsegment is parallel to the third side and half as long.
Worked example: Finding lengths
$D$ and $E$ are the midpoints of $\overline{AB}$ and $\overline{AC}$, and $BC = 26$. Find $DE$. If $DE$ were $9.5$, what would $BC$ be?
- $DE = \frac{1}{2} \cdot 26 = 13$.
- Going the other way, the third side is twice the midsegment: $BC = 2 \cdot 9.5 = 19$.
Answer: $DE = 13$; $BC = 19$
Worked example: With algebra
$\overline{DE}$ is a midsegment of $\triangle ABC$, parallel to $\overline{BC}$, with $DE = x + 4$ and $BC = 3x - 2$. Find $BC$.
- $BC = 2 \cdot DE$: $3x - 2 = 2(x + 4) = 2x + 8$.
- $x = 10$, so $BC = 3(10) - 2 = 28$ (and $DE = 14$).
Answer: $BC = 28$
Worked example: Checking with coordinates
$A(2, 6)$, $B(0, 0)$ and $C(8, 2)$. $D$ and $E$ are the midpoints of $\overline{AB}$ and $\overline{AC}$. Verify the theorem for $\overline{DE}$.
- $D = \left(\frac{2 + 0}{2}, \frac{6 + 0}{2}\right) = (1, 3)$ and $E = \left(\frac{2 + 8}{2}, \frac{6 + 2}{2}\right) = (5, 4)$.
- Slopes: $\overline{DE}$ has $\frac{4 - 3}{5 - 1} = \frac{1}{4}$ and $\overline{BC}$ has $\frac{2 - 0}{8 - 0} = \frac{1}{4}$. Equal slopes: parallel.
- Lengths: $DE = \sqrt{4^2 + 1^2} = \sqrt{17}$ and $BC = \sqrt{8^2 + 2^2} = \sqrt{68} = 2\sqrt{17}$. Half as long.
Answer: $\overline{DE} \parallel \overline{BC}$ and $DE = \frac{1}{2}BC$, as promised.
Common mistake
Pairing a midsegment with the wrong side. The midsegment is parallel to (and half of) the side it does not touch. Joining the midpoints of $\overline{AB}$ and $\overline{AC}$ gives a segment parallel to $\overline{BC}$. Check: its endpoints sit on the two other sides.