## Centre of Triangle Using Vectors

We can prove that the centre of a triangle is two thirds of the way from a vertex to the midpoint of the opposite side with vectors.
Consider the triangle below, with  {jatex options:inline}\mathbf{AB}=\mathbf{b}, \; \mathbf{AC}= \mathbf{c}{/jatex}  and  {jatex options:inline}P, \; Q{/jatex}  as the midpoints of  {jatex options:inline}BC, \; AC{/jatex}  respectively.

We need  {jatex options:inline}\mathbf{BC}= - \mathbf{b}+ \mathbf{c} \rightarrow \mathbf{BP}= \frac{1}{2}(- \mathbf{b}+ \mathbf{c}){/jatex}.
The equation of the line  {jatex options:inline}BQ{/jatex}  is  {jatex options:inline}\mathbf{b}+ (- \mathbf{b}+ \frac{\mathbf{c}})s=(1-s) \mathbf{b}+ \frac{s}{2} \mathbf{c}{/jatex}.
The equation of the line  {jatex options:inline}AP{/jatex}  is  {jatex options:inline}(\mathbf{b} + \mathbf{c}{2})t= \frac{t}{2} \mathbf{b} + \frac{t}{2} \mathbf{c}{/jatex}  (remembering that  {jatex options:inline}\mathbf{b}, \; \mathbf{c}{/jatex}  form a parallelogram, of which the triangle forms a half.
Equating the coefficients of  {jatex options:inline}\mathbf{b}, \; \mathbf{c}{/jatex}  to find the intersection gives {jatex options:inline}1-s= \frac{t}{2}{/jatex}
{jatex options:inline}\frac{s}{2}= \frac{t}{2}{/jatex}
From the second  {jatex options:inline}s=t{/jatex}  and from the first  {jatex options:inline}s=t= \frac{2}{3}{/jatex}.