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Question: In a triangle, a median is the line segment that connects a vertex with the midpoint of the opposite side. Prove that in a triangle with side lengths a, b, and c, the length Mc of the median, drawn to the side with the length c, is equal to Mc = 1/2 * √(2*a^2+2*b^2-c^2)

I have been hinted that I can begin by finding the cosine of the angle opposite to side b. I thought about beginning by trying to find the area of the triangle, but I am not sure if that would work and how I should proceed.

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marked as duplicate by dxiv, Henrik, user91500, Especially Lime, Ken Duna Apr 27 '17 at 14:15

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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.

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