show that the medians of a triangle ABC are coincident
Let $\displaystyle \overrightarrow a,\overrightarrow b,\overrightarrow c$, denote the vertices $\displaystyle A,B,C$ respectively and also assume $\displaystyle D(\overrightarrow d),E(\overrightarrow e),F(\overrightarrow f)$ denote the mid-points of $\displaystyle AB,BC,CA$ respectively.Therefore,
$\displaystyle \overrightarrow d=\frac{\overrightarrow b+\overrightarrow c}{2}$
$\displaystyle \overrightarrow e=\frac{\overrightarrow c+\overrightarrow a}{2}$
$\displaystyle \overrightarrow f=\frac{\overrightarrow a+\overrightarrow b}{2}$
$\displaystyle
2\overrightarrow d=\overrightarrow b+\overrightarrow c;
2\overrightarrow e=\overrightarrow c+\overrightarrow a;
2\overrightarrow f=\overrightarrow a+\overrightarrow b
$
$\displaystyle 2\overrightarrow d+\overrightarrow a=2\overrightarrow e+\overrightarrow b=2\overrightarrow f+\overrightarrow c=\overrightarrow a+\overrightarrow b+\overrightarrow c$
$\displaystyle
\frac{2\overrightarrow d+\overrightarrow a}{3}=
\frac{2\overrightarrow e+\overrightarrow b}{3}=
\frac{2\overrightarrow f+\overrightarrow c}{3}=\frac{\overrightarrow a+\overrightarrow b+\overrightarrow c}{3}
$
Now,$\displaystyle \frac{2\overrightarrow d+\overrightarrow a}{3}$ is the position vector of the point on median $\displaystyle AD$ which divides $\displaystyle AD$ in the ratio $\displaystyle 2:1$.Simlarly,$\displaystyle \frac{2\overrightarrow e+\overrightarrow b}{3}$ is the position vector of the point on median $\displaystyle BE$ which divides $\displaystyle BE$ in the ratio $\displaystyle 2:1$ and $\displaystyle \frac{2\overrightarrow f+\overrightarrow c}{3}$ is the position vector of the point on median $\displaystyle CF$ which divides $\displaystyle CF$ in the ratio $\displaystyle 2:1$.
Since these position vectors have turned out to be equal,obviously $\displaystyle AD,BE\\,and\\,CF$ are concurrent and the point of concurrency is $\displaystyle \frac{\overrightarrow a+\overrightarrow b+\overrightarrow c}{3}$