# Find the length of the median of a triangle using three side lengths [duplicate]

Let $C'$ be the midpoint of $AB$ and $AB=c,\, AC=b,\, BC=a$ as usual.
By the cosine theorem $$\cos\widehat{BAC} = \frac{b^2+c^2-a^2}{2bc}$$ as well as $$m_c^2 = CC'^2 = AC^2+AC'^2- 2 AC\cdot AC'\cos\widehat{BAC}$$ from which $$m_c^2 = b^2+\frac{c^2}{4}- bc\cdot\frac{b^2+c^2-a^2}{2bc} = \frac{2a^2+2b^2-c^2}{4}$$ and $m_c=\frac{1}{2}\sqrt{2a^2+2b^2-c^2}$ as wanted.