The Midsegment of a Triangle

In a triangle ABC if M is the middle of AB and N is the middle of CA then MN is called the midsegment of the triangle.

MN is parallel with BC and is also half the size of BC.

Proof:
Let O be the symmetric of M in respect to N.

N is the middle of AC and the middle of MO. This means that AMCO is a parallelogram. So $AM \parallel OC$ and $AM = OC$ .

But because M is the middle of BC we have that $AM = MB$ . So $MB = OC$ .
$AM \parallel OC$ and A, M, B collinear means that $MB \parallel OC$ .

$MB = OC$ and $MB \parallel OC$ means that MBCO is a parallelogram.
MBCO is a parallelogram means that $BC = MO$ and $BC \parallel MO$ . Because N is the middle of MO we have:
$MN = \frac{BC}{2}$ and $MN \parallel BC$