Median Formula/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 $\mathbf a = \overrightarrow {AC}$ and $\mathbf b = \overrightarrow {BC}$.

Then:

\(\ds \overrightarrow {AB}\)

\(=\)

\(\ds \mathbf a - \mathbf b\)

\(\ds \overrightarrow {AD}\)

\(=\)

\(\ds \dfrac {\overrightarrow {AB} } 2\)

\(\ds \)

\(=\)

\(\ds \dfrac {\mathbf a - \mathbf b} 2\)

Then:

\(\ds \overrightarrow {AC} + \overrightarrow {CD}\)

\(=\)

\(\ds \overrightarrow {AD}\)

\(\ds \leadsto \ \ \)

\(\ds \overrightarrow {CD}\)

\(=\)

\(\ds \overrightarrow {AD} - \overrightarrow {AC}\)

\(\ds \)

\(=\)

\(\ds \frac {\mathbf a - \mathbf b} 2 - \mathbf a\)

\(\ds \)

\(=\)

\(\ds -\frac {\mathbf a + \mathbf b} 2\)

\(\ds \leadsto \ \ \)

\(\ds {m_c}^2\)

\(=\)

\(\ds \size {\overrightarrow {CD} }^2\)

\(\ds \)

\(=\)

\(\ds \size {-\frac {\mathbf a + \mathbf b} 2}^2\)

\(\ds \)

\(=\)

\(\ds \frac 1 4 \paren {\paren {\mathbf a + \mathbf b} \cdot \paren {\mathbf a + \mathbf b} }\)

Dot Product of Vector with Itself

\(\ds \)

\(=\)

\(\ds \frac 1 4 \paren {\mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b + 2 \mathbf a \cdot \mathbf b}\)

Square of Sum of Vectors

\(\ds \)

\(=\)

\(\ds \frac 1 4 \paren {\mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b - \paren {\mathbf a - \mathbf b} \cdot \paren {\mathbf a - \mathbf b} + \mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b}\)

Square of Sum of Vectors

\(\ds \)

\(=\)

\(\ds \frac 1 4 \paren {2 \size {\mathbf a}^2 + 2 \size {\mathbf b}^2 - \size {\mathbf a - \mathbf b}^2}\)

Dot Product of Vector with Itself

\(\ds \)

\(=\)

\(\ds \frac 1 4 \paren {2 a^2 + 2 b^2 - c^2}\)

\(\ds \)

\(=\)

\(\ds \frac {a^2 + b^2} 2 - \frac {c^2} 4\)

$\blacksquare$