Prove by vector method, medians of a triangle are concurrent.

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Solution

AD,BE and CF are medians
Let $O$ be the origin and let
$\to O A = \to a, \to O B = \to b, \to O C = \to c, \to O D = \to d, \to O E = \to e, \to O F = \to f$ be th posiotion vectors points $A, B, C, D, E, F$ respectively
From mid-point formula, we get
$\to d = \frac{\to b + \to c}{2}$
$2 \to d = \to b + \to c$
On adding $\to a$ and then dividing by 3 both sides
$\frac{2 \to d + \to a}{3} = \frac{\to a + \to b + \to c}{3}$

$\to e = \frac{\to a + \to c}{2}$
$2 \to e = \to a + \to c$
On adding $\to b$ and then dividing by 3 both sides
$\frac{2 \to e + \to b}{3} = \frac{\to a + \to b + \to c}{3}$

$\to f = \frac{\to a + \to b}{2}$
$2 \to f = \to a + \to b$
On adding $\to c$ and then dividing by 3 both sides
$\frac{2 \to f + \to c}{3} = \frac{\to a + \to b + \to c}{3}$

But $\frac{\to a + \to b + \to c}{3}$ is the centroid of traingle ABC
SO the point represented by $\frac{\to a + \to b + \to c}{3}$ lies on medians AD,BE,CF

Hence medians are concurrent

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