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, 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

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