Question

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

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

Answer 1

Let A, B and C be vertices of a triangle.

Let D, E and F be the mid-points of the sides BC, AC and AB respectively.

Let ${\displaystyle \overline{a},\overline{b},\overline{c},\overline{d},\overline{e}}$ and ${\displaystyle \overline{f}}$ be position vectors of points A, B, C, D, E and F respectively.

Therefore, by mid-point formula,

∴ ${\displaystyle \overline{d}=\frac{\overline{b}+\overline{c}}{2},\overline{e}=\frac{\overline{a}+\overline{c}}{2}}$ and ${\displaystyle \overline{f}=\frac{\overline{a}+\overline{b}}{2}}$

∴ ${\displaystyle 2\overline{d}=\overline{b}+\overline{c},2\overline{e}=\overline{a}+\overline{c}}$ and ${\displaystyle 2\overline{f}=\overline{a}+\overline{b}}$

∴ ${\displaystyle 2\overline{d}+\overline{a}=\overline{a}+\overline{b}+\overline{c}}$, similarly ${\displaystyle 2\overline{e}+\overline{b}=2\overline{f}+\overline{c}=\overline{a}+\overline{b}+\overline{c}}$

∴ ${\displaystyle \frac{2\overline{d}+\overline{a}}{3}=\frac{2\overline{e}+\overline{b}}{3}=\frac{2\overline{f}+\overline{c}}{3}=\frac{\overline{a}+\overline{b}+\overline{c}}{3}=\overline{g}}$  ...(Say)

Then we have ${\displaystyle \overline{g}=\frac{\overline{a}+\overline{b}+\overline{c}}{3}=\frac{\left(2\right)\overline{d}+\left(1\right)\overline{a}}{2+1}=\frac{\left(2\right)\overline{e}+\left(1\right)\overline{b}}{2+1}=\frac{\left(2\right)\overline{f}+\left(1\right)\overline{c}}{2+1}}$

If G is the point whose position vector is ${\displaystyle \overline{g}}$, then from the above equation it is clear that the point G lies on the medians AD, BE, CF and it divides each of the medians AD, BE, CF internally in the ratio 2 : 1.

Therefore, three medians are concurrent.