Medians of Triangle Meet at Centroid/Proof 2
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Theorem
Let $\triangle ABC$ be a triangle.
Then the medians of $\triangle ABC$ meet at a single point.
This point is called the centroid of $\triangle ABC$.
Proof
Let $\vec a, \vec b, \vec c$ be $\vec{OA}, \vec{OB}, \vec{OC}$ respectively.
Let the midpoint of $BC, AC, AB$ be $\vec d, \vec e, \vec f$ respectively.
Then:
\(\ds \vec d\) | \(=\) | \(\ds \frac {\vec b + \vec c} 2\) | ||||||||||||
\(\ds \vec e\) | \(=\) | \(\ds \frac {\vec a + \vec c} 2\) | ||||||||||||
\(\ds \vec f\) | \(=\) | \(\ds \frac {\vec a + \vec b} 2\) |
The three medians are $\vec{AD}, \vec{BE}, \vec{CF}$ respectively:
\(\ds \vec {AD}\) | \(=\) | \(\ds \vec d - \vec a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\vec b + \vec c - 2 \vec a} 2\) | ||||||||||||
\(\ds \vec {BE}\) | \(=\) | \(\ds \vec e - \vec b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\vec a + \vec c - 2 \vec b} 2\) | ||||||||||||
\(\ds \vec {CF}\) | \(=\) | \(\ds \vec f - \vec c\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\vec a + \vec b - 2 \vec c} 2\) |
Their equations:
\(\text {(1)}: \quad\) | \(\ds \vec {AD}: \ \ \) | \(\ds \vec r\) | \(=\) | \(\ds \vec a + x \paren {\frac {\vec b + \vec c - 2\vec a} 2}\) | ||||||||||
\(\text {(2)}: \quad\) | \(\ds \vec {BE}: \ \ \) | \(\ds \vec r\) | \(=\) | \(\ds \vec b + y \paren {\frac {\vec a + \vec c - 2\vec b} 2}\) | ||||||||||
\(\text {(3)}: \quad\) | \(\ds \vec {CF}: \ \ \) | \(\ds \vec r\) | \(=\) | \(\ds \vec c + z \paren {\frac {\vec a + \vec b - 2\vec c} 2}\) |
It can be verified that $x = y = z = \dfrac 2 3$ produce the same point:
When $x = \dfrac 2 3$, from $(1)$:
- $\vec r = \vec a + \dfrac 2 3 \paren {\dfrac {\vec b + \vec c - 2\vec a} 2} = \dfrac {\vec a + \vec b + \vec c} 3$
When $y = \dfrac 2 3$, from $(2)$:
- $\vec r = \vec b + \dfrac 2 3 \paren {\dfrac {\vec a + \vec c - 2\vec b} 2} = \dfrac {\vec a + \vec b + \vec c} 3$
When $z = \dfrac 2 3$, from $(3)$:
- $\vec r = \vec c + \dfrac 2 3 \paren {\dfrac {\vec a + \vec b - 2\vec c} 2} = \dfrac {\vec a + \vec b + \vec c} 3$
Therefore, the three medians meet at a single point, namely $\dfrac {\vec a + \vec b + \vec c} 3$.
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $67$