If ${A,B,C}$ be the position vectors of the vertices A,B,C of the triangle ${ABC}$,show that the three medians concur at the point ${\frac{1}{3}( A + B + C)}$, called the centroid.
Note : I don't need the solution. I just need to know how to prove the concurrence of two or more vectors. Thank you! :)
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Let $M_A=\frac{B+C}{2},M_B=\frac{A+C}{2},M_C=\frac{A+B}{2}$ and $G=\frac{A+B+C}{3}$.
You can check that $$ 3 G = 2 M_A + A,\tag{1}$$ hence $G$ lies on the $AM_A$ line and satisfies $\frac{GA}{GM_A}=2$, but the same holds if in $(1)$ you replace $A$ with $B$ or $C$, hence all the medians concur in $G$. The standard tool to prove concurrencies in a triangle is Ceva's theorem, in its metric or trigonometric form.
Suppose the vertices $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$
Then the midpoint of $BC$ will be $D(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$
the midpoint of $AC$ will be $E(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})$
the midpoint of $AB$ will be $F(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
Now you can find the equation of lines containing the medians $AD,BE,CF$
Solve this system of equations,you will get the intesection point $M=\frac{A+B+C}{3}$