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, $latex PQR$ is a triangle and $latex MN$ 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**

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

**Proof**

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

Since $latex M$ is the midpoint of $latex PQ$, by the midpoint formula,

$latex M = (\displaystyle\frac{2b+0}{2} ,\displaystyle\frac{ 2c + 0}{2}) = (b,c)$.

Now, $latex N$ is also the midpoint of $latex PR$, so

$latex N = (\displaystyle\frac{2b+2a}{2},\displaystyle\frac{2c + 0}{2}) = (b + a , c)$.

It follows that the slope of

$latex MN = \displaystyle\frac{c-c}{b + a – b} = 0$.

and the slope of

$latex QR = \displaystyle\frac{0-0}{2a-0} =0$.

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

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

By the distance formula,

$latex QR = \sqrt{(2a-0)^2 + (0-0)^2} = \sqrt{4a^2} = 2 \sqrt{a^2}$.

For the length of $latex MN$, we have

$latex MN = \sqrt{((b+a)-b)^2 + (c-c)^2} = \sqrt{a^2}$

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

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

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

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