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The Midsegment of a Triangle

Posted November 17th, 2007 by Isoscel

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 $

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Comments

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