Median length theorem

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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