Median Formula/Proof 1

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:

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

$\ds a^2 \cdot AD + b^2 \cdot DB$

$=$

$\ds CD^2 \cdot c + AD \cdot DB \cdot c$

$\ds \leadsto \ \ $

$\ds a^2 \frac c 2 + b^2 \frac c 2$

$=$

$\ds {m_c}^2 \cdot c + \paren {\frac c 2}^2 c$

substituting $AD = DB = \dfrac c 2$ and $CD = m_c$

$\ds \leadsto \ \ $

$\ds \frac c 2 \paren {a^2 + b^2}$

$=$

$\ds m_c^2 \cdot c + \frac {c^2} 4 \cdot c$

$\ds \leadsto \ \ $

$\ds \frac {a^2 + b^2} 2$

$=$

$\ds m_c^2 + \frac {c^2} 4$

$\ds \leadsto \ \ $

$\ds {m_c}^2$

$=$

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

after algebra

$\blacksquare$