# The Proof of the Midsegment Theorem

The segment connecting the two midpoints of the sides of a triangle is called its midsegment or midline. The midsegment of a triangle has interesting characteristics:

(1) the midsegment connecting the midpoints of the two sides of a triangle is parallel to the third side and (2) its length is also half of the third side. In this post, we prove these two theorems. In the given above, $PQR$ is a triangle and $MN$ is a midsegment.

Theorem

The segment connecting the two sides of a triangle is parallel to and half the length of the third side.

What To Show

1. $MN$ is parallel to $QR$
2. $MN = \frac{1}{2}QR$.

Proof

Let $PQR$ be  a triangle with points $P$, $Q$, and $R$ having coordinates $(0,0)$, $Q(2b, 2c)$ and $(2a,0)$ respectively as shown in the figure above.

Since $M$ is the midpoint of $PQ$, by the midpoint formula, $M = (\displaystyle\frac{2b+0}{2} ,\displaystyle\frac{ 2c + 0}{2}) = (b,c)$.

Now, $N$ is also the midpoint of $PR$, so $N = (\displaystyle\frac{2b+2a}{2},\displaystyle\frac{2c + 0}{2}) = (b + a , c)$.

It follows that the slope of $MN = \displaystyle\frac{c-c}{b + a - b} = 0$.

and the slope of $QR = \displaystyle\frac{0-0}{2a-0} =0$.

Since $MN$ and $QR$ have the same slope, they are parallel. We have proved the first theorem.

Now, it remains to show that $QR$ is half that of $MN$.

By the distance formula, $QR = \sqrt{(2a-0)^2 + (0-0)^2} = \sqrt{4a^2} = 2 \sqrt{a^2}$.

For the length of $MN$, we have $MN = \sqrt{((b+a)-b)^2 + (c-c)^2} = \sqrt{a^2}$

So, $MN = \frac{1}{2} QR$

## 3 thoughts on “The Proof of the Midsegment Theorem”

1. Shaun on said:

I didn’t know about these midsegment theorems, but you have done a good job demonstrating the proof!

• Guillermo Bautista on said:

Thanks Shaun. You have probably forgotten it since it is taught in high school. 🙂