Median Formula
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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 1
| \(\ds a^2 \cdot AD + b^2 \cdot DB\) | \(=\) | \(\ds CD^2 \cdot c + AD \cdot DB \cdot c\) | Stewart's Theorem | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a^2 \frac c 2 + b^2 \frac c 2\) | \(=\) | \(\ds {m_c}^2 \cdot c + \paren {\frac c 2}^2 c\) | 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 |
$\blacksquare$
Proof 2
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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {2 a^2 + 2 b^2 - c^2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a^2 + b^2} 2 - \frac {c^2} 4\) |
$\blacksquare$
Proof 3
| \(\ds {m_c}^2\) | \(=\) | \(\ds b^2 + \paren {\frac c 2}^2 - 2 b \paren {\frac c 2} \cos A\) | 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 | |||||||||||
| \(\ds \) | \(=\) | \(\ds b^2 + \frac {c^2} 4 - \frac {b^2 + c^2 - a^2} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a^2} 2 + \frac {b^2} 2 - \frac {c^2} 4\) |
$\blacksquare$
Examples
Triangle $\tuple {1, -2}, \tuple {-3, 4}, \tuple {2, 2}$
Consider the triangle $\triangle ABC$ whose vertices are:
- $A = \tuple {1, -2}, B = \tuple {-3, 4}, C = \tuple {2, 2}$
The length of the median of $\triangle ABC$ which which bisects $AB$ is $\sqrt {10}$.
Sources
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): median formula