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  
\[\mathbf{AB}=\mathbf{b}, \; \mathbf{AC}= \mathbf{c}\]
  and  
\[P, \; Q\]
  as the midpoints of  
\[BC, \; AC\]
  respectively.

center of triangle

We need  
\[\mathbf{BC}= - \mathbf{b}+ \mathbf{c} \rightarrow \mathbf{BP}= \frac{1}{2}(- \mathbf{b}+ \mathbf{c})\]
.
The equation of the line  
\[BQ\]
  is  
\[\mathbf{b}+ (- \mathbf{b}+ \frac{\mathbf{c}})s=(1-s) \mathbf{b}+ \frac{s}{2} \mathbf{c}\]
.
The equation of the line  
\[AP\]
  is  
\[(\mathbf{b} + \mathbf{c}{2})t= \frac{t}{2} \mathbf{b} + \frac{t}{2} \mathbf{c}\]
  (remembering that  
\[\mathbf{b}, \; \mathbf{c}\]
  form a parallelogram, of which the triangle forms a half.
Equating the coefficients of  
\[\mathbf{b}, \; \mathbf{c}\]
  to find the intersection gives
\[1-s= \frac{t}{2}\]

\[\frac{s}{2}= \frac{t}{2}\]

From the second  
\[s=t\]
  and from the first  
\[s=t= \frac{2}{3}\]
.

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