Triangle/Medians/Intersecting point/Fact/Proof2

Proof

We consider the condition

$\frac{B+C}{2}+s(A-\frac{B+C}{2})=\frac{A+C}{2}+t(B-\frac{A+C}{2}),$

which yields

$\begin{array}{cc}\frac{B-A}{2}& =t(B-\frac{A+C}{2})-s(A-\frac{B+C}{2})\\ & =(-s-\frac{t}{2})A+(t+\frac{s}{2})B+\frac{s-t}{2}C.\end{array}$

We may assume $E={ℝ}^2$ and $A=0$. This implies, because $B$ and $C$ are linearly independent, the identity

$s=t.$

Therefore,

$\frac{B-A}{2}=(-\frac{3}{2}s)A+\left(\frac{3}{2}s\right)B,$

and we obtain

$s=\frac{1}{3}.$

Hence, the intersecting point is

$\frac{B+C}{2}+\frac{1}{3}(A-\frac{B+C}{2})=\frac{1}{3}A+\frac{1}{3}B+\frac{1}{3}C.$

Due to the symmetry of the situation, this is also the intersecting point with the third median.

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