Length of Median of Triangle

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 1

 $\displaystyle a^2 \cdot AD + b^2 \cdot DB$ $=$ $\displaystyle CD^2 \cdot c + AD \cdot DB \cdot c$ Stewart's Theorem $\displaystyle \leadsto \ \$ $\displaystyle a^2 \frac c 2 + b^2 \frac c 2$ $=$ $\displaystyle {m_c}^2 \cdot c + \left({\frac c 2}\right)^2 c$ substituting $AD = DB = \dfrac c 2$ and $CD = m_c$ $\displaystyle \leadsto \ \$ $\displaystyle \frac c 2 \left({a^2 + b^2}\right)$ $=$ $\displaystyle m_c^2 \cdot c + \frac {c^2} 4 \cdot c$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {a^2 + b^2} 2$ $=$ $\displaystyle m_c^2 + \frac {c^2} 4$ $\displaystyle \leadsto \ \$ $\displaystyle {m_c}^2$ $=$ $\displaystyle \frac {a^2 + b^2} 2 - \frac {c^2} 4$ after algebra

$\blacksquare$

Proof 2

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

Examples

Triangle $\paren {1, -2}, \paren {-3, 4}, \paren {2, 2}$

Consider the triangle $\triangle ABC$ whose vertices are:

$A = \paren {1, -2}, B = \paren {-3, 4}, C = \paren {2, 2}$

The length of the median of $\triangle ABC$ which which bisects $AB$ is $\sqrt {10}$.