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

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

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”

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

Reply

Guillermo Bautista on 28 March, 2013 at 2:28 pm said:

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

Pingback: The Midpoints of the Sides of a Quadrilateral Form a Parallelogram


Shaun on 28 March, 2013 at 1:59 pm said:

I didn’t know about these midsegment theorems, but you have done a good job demonstrating the proof!
- Reply
  
  Guillermo Bautista on 28 March, 2013 at 2:28 pm said:
  
  Thanks Shaun. You have probably forgotten it since it is taught in high school. 🙂
Pingback: The Midpoints of the Sides of a Quadrilateral Form a Parallelogram

3 thoughts on “The Proof of the Midsegment Theorem”

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