By vector method prove that the medians of a triangle are concurrent

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Solution

Let $A, B$ and $C$ be vertices of a triangle.
Let $D, E$ and $F$ be the midpoints of the sides $B C, A C$ and $A B$ respectively. Let $\to O A = \to a, \to O B = \to b, \to O C = \to c, \to O D = \to d = \to O E = \to e$ and $\to O F = \to f$ be position vectors of points $A, B, C, D, E$ and $F$ respectively.
Therefore, by Midpoint formula,
$\to d = \frac{\to b + \to c}{2}, \to e = \frac{\to a + \to c}{2}$ and $\to f = \frac{\to a + \to b}{2}$
$∴ 2 \to d = \to b + \to c, 2 \to e = \to a + \to c$ and $2 \to f = \to a + \to b$
$∴ 2 \to d + \to a = \to a + \to b + \to c$ ,
$2 \to e + \to b = \to a + \to b + \to c$ ,
$2 \to e + \to b = \to a + \to b + \to c$
$2 \to f + \to c = \to a + \to b + \to c$
Now, $\frac{2 \to d + \to a}{3} = \frac{2 \to e + \to b}{3} = \frac{2 \to f + \to c}{3} = \frac{\to a + \to b + \to c}{3}$
Let $\to g = \frac{\to a + \to b + \to c}{3}$ . Then, we have
$\to g = \frac{\to a + \to b + \to c}{3} = \frac{(2) \to d + (1) \to a}{2 + 1}$
$= \frac{(2) \to e + (1) \to b}{2 + 1} = \frac{(2) \to f + (1) \to c}{2 + 1}$
If $G$ is the point whose position vector is $\to g$ , then from the above equation it is clear that the point $G$ lies on the medians $A B, B E, C F$ and it divides them internally in the ratio $2 : 1$ .
Hence, the medians of a triangle are concurrent.

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