Median Formula/Proof 3
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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
\(\ds {m_c}^2\) | \(=\) | \(\ds b^2 + \paren {\frac c 2}^2 - 2 b \paren {\frac c 2} \cos A\) | Law of Cosines | |||||||||||
\(\ds \) | \(=\) | \(\ds b^2 + \frac {c^2} 4 - 2 b \paren {\frac c 2} \paren {\frac {b^2 + c^2 - a^2} {2 b c} }\) | Law of Cosines | |||||||||||
\(\ds \) | \(=\) | \(\ds b^2 + \frac {c^2} 4 - \frac {b^2 + c^2 - a^2} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a^2} 2 + \frac {b^2} 2 - \frac {c^2} 4\) |
$\blacksquare$