# Length of Median of Triangle/Proof 2

## Theorem

Let $\triangle ABC$ be a triangle.

Let $CD$ be the median of $\triangle ABC$ which bisects $AB$.

The length $m_c$ of $CD$ is given by:

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

## Proof

Let $\triangle ABC$ be embedded in the complex plane.

Let $A = \tuple {x_1, y_1}$, $B = \tuple {x_2, y_2}$ and $C = \tuple {x_3, y_3}$ be represented by the complex numbers $z_1$, $z_2$ and $z_3$ respectively.

Then:

 $\displaystyle AC$ $=$ $\displaystyle z_3 - z_1$ $\displaystyle BC$ $=$ $\displaystyle z_3 - z_2$ $\displaystyle AB$ $=$ $\displaystyle z_2 - z_1$ $\displaystyle AD$ $=$ $\displaystyle \dfrac {AB} 2$ $\displaystyle$ $=$ $\displaystyle \paren {\dfrac {z_2 - z_1} 2}$

Then:

 $\displaystyle AC + CD$ $=$ $\displaystyle AD$ $\displaystyle \leadsto \ \$ $\displaystyle CD$ $=$ $\displaystyle AD - AC$ $\displaystyle$ $=$ $\displaystyle \paren {\dfrac {z_2 - z_1} 2} - \paren {z_3 - z_1}$ $\displaystyle$ $=$ $\displaystyle \paren {\dfrac {z_1 - z_3} 2} + \paren {\dfrac {z_2 - z_3} 2}$