Length of Median of Triangle/Proof 2

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Theorem

Let $\triangle ABC$ be a triangle.

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

MedianOfTriangle.png

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.


Length-of-Triangle-Median-Complex.png


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:

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