Using vector method prove that the medians of a triangle are concurrent.

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Solution

Let $\to a, \to b, \to c$ be the position vectors of the vertices $A, B, C$ of $Δ A B C$ and $\to d, \to e, \to f$ be the position vectors of the midpoints $D, E, F$ of the sides $B C, C A, A B$ respectively.

Then by the midpoint formula, $\to d = \frac{\to b + \to c}{2}, \to e = \frac{\to c + \to a}{2}, \to f = \frac{\to a + \to b}{2}$

$∴ 2 \to d = \to b + \to c; 2 \to e = \to c + \to a; 2 \to f = \to a + \to b$

$∴ 2 \to d + \to a = \to a + \to b + \to c, 2 \to e + \to b = \to a + \to b + \to c, 2 \to f + \to c = \to a + \to b + \to c$

$∴ \frac{2 \to d + \to a}{2 + 1} = \frac{2 \to e + \to b}{2 + 1} = \frac{2 \to f + \to c}{2 + 1} = \frac{\to a + \to b + \to c}{3} = \to g (s a y)$ lies on the three medians $A D, B E, C F$ dividing each of them internally in the ratio $2 : 1$ .

Hence the medians are concurrent at point $G$ .

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