Triangle/Medians/Intersecting point/Fact/Proof2

We consider the condition

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

which yields

${\displaystyle {}{\begin{aligned}{\frac {B-A}{2}}&=t{\left(B-{\frac {A+C}{2}}\right)}-s{\left(A-{\frac {B+C}{2}}\right)}\\&={\left(-s-{\frac {t}{2}}\right)}A+{\left(t+{\frac {s}{2}}\right)}B+{\frac {s-t}{2}}C.\end{aligned}}}$

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

${\displaystyle {}s=t\,.}$

Therefore,

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

and we obtain

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

Hence, the intersecting point is

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

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