Median length theorem

Submitted by Structure on Sat, 12/29/2007 - 18:25.

Let ABC a triangle and a=length(BC),b=length(CA),c=length(AB).
Let M be the middle of BC and D symmetric of A with respect of M.
Then ABDC is a parallelogram.

From cosine theorem we have

$a^2=b^2+c^2-2bc \cos A$

In triangle ABD we have

$4m_a^2=AD^2=b^2+c^2-2bc\cos(\pi-A)$

Summing up we have

$a^2+4m_a^2=2b^2+2c^2-2bc(\cos A+\cos(\pi-A))=2b^2+2c^2$

Finally

$m_a^2=\frac{1}{4}(2(b^2+c^2)-a^2)$

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Median length theorem