Triangle/Medians/Intersecting point/Fact/Proof

We have

${\displaystyle {}{\frac {B+C}{2}}+{\frac {1}{3}}{\left(A-{\frac {B+C}{2}}\right)}={\frac {1}{3}}A+{\left({\frac {1}{2}}-{\frac {1}{6}}\right)}B+{\left({\frac {1}{2}}-{\frac {1}{6}}\right)}C={\frac {1}{3}}A+{\frac {1}{3}}B+{\frac {1}{3}}C\,,}$

and this is a point of the median through ${\displaystyle {}A}$. Due to the symmetry of the situation, this point also lies on the other medians.