Length of Median

Theorem

The length of a median is equal to

$\displaystyle m_a^2=\dfrac{c^2}{2}+\dfrac{b^2}{2}-\dfrac{a^2}{4}$

$\displaystyle m_b^2=\dfrac{c^2}{2}+\dfrac{a^2}{2}-\dfrac{b^2}{4}$

$\displaystyle m_c^2=\dfrac{a^2}{2}+\dfrac{b^2}{2}-\dfrac{c^2}{4}$

Where $a$, $b$, and $c$ are the sides opposite $A$, $B$, and $C$ respectively.

And $m_a$, $m_b$, and $m_c$ are the medians from $A$, $B$, and $C$ respectively.

Proof

We look at one of the medians, WLOG $m_c$:

We use Stewart's Theorem, noting that $\displaystyle AP=PB=\frac{c}{2}$ and $CP = m_c$

From Stewart's Theorem, this gives us

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{c}{2}(a^2+b^2)\)	\(=\)	\(\displaystyle m_c^2 \cdot c+\frac{c^2}{4}\cdot c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{a^2+b^2}{2}\)	\(=\)	\(\displaystyle m_c^2+\frac{c^2}{4}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle m_c^2\)	\(=\)	\(\displaystyle \frac{a^2+b^2}{2}-\frac{c^2}{4}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

A similar argument can be used to show that the statement holds for the others medians.