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.

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
$s=t= \frac{2}{3}$
.