Question
Using vector method, prove that the median of a triangle are concurrent.
✓
Answer
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 $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}$ and $\bar{f}$ be position vectors of points $A , B , C , D , E$ and $F$ respectively.
Therefore, by mid-point formula,
$\therefore \bar{d}=\frac{\bar{b}+\bar{c}}{2}, \bar{e}=\frac{\bar{a}+\bar{c}}{2}$ and $\bar{f}=\frac{\bar{a}+\bar{b}}{2}$
$\therefore 2 \bar{d}=\bar{b}+\bar{c}, 2 \bar{e}=\bar{a}+\bar{c}$ and $2 \bar{f}=\bar{a}+\bar{b}$
$\therefore 2 \bar{d}+\bar{a}=\bar{a}+\bar{b}+\bar{c}$, similarly $2 \bar{e}+\bar{b}=2 \bar{f}+\bar{c}=\bar{a}+\bar{b}+\bar{c}$
$\therefore \frac{2 \bar{d}+\bar{a}}{3}=\frac{2 \bar{e}+\bar{b}}{3}=\frac{2 \bar{f}+\bar{c}}{3}=\frac{\bar{a}+\bar{b}+\bar{c}}{3}=\bar{g} \ldots$ (Say)
Then we have
$\bar{g}=\frac{\bar{a}+\bar{b}+\bar{c}}{3}=\frac{(2) \bar{d}+(1) \bar{a}}{2+1}=\frac{(2) \bar{e}+(1) \bar{b}}{2+1}=\frac{(2) \bar{f}+(1) \bar{c}}{2+1}$
If G is the point whose position vector is $\bar{g}$,
Therefore, three medians are concurrent.
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