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.

midsegment

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

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

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

3 thoughts on “The Proof of the Midsegment Theorem

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

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