Triangle property intersection proof with vectors

Question

Given that:

$\vec{OA}=a$

$\vec{OB}=b$

$\vec{OC}=c$

And that point R and F are the mid-points of $\vec{AC}$ and $\vec{BC}$ respectively, show that $\vec{OG}=\frac{1}{3}(a+b+c)$. I am really stuck with this question. What I tried is to show that $\kappa\vec{AF}=\lambda\vec{RB}$ where $\lambda$ and $\kappa$ are constants, but It gets more complicated. Any ideas?

Answer 1

$\vec{AC}=c-a$

$\vec{AR}={1\over2}(c-a)$

$\vec{OR}={1\over2}(c-a)+a={1\over2}(c+a)$

$\vec{AB}=b-a$

$\vec{RB}=(b-a)-{1\over2}(c-a)=b-{1\over2}(c+a)$

$\vec{RG}={1\over3}(b-{1\over2}(c+a))$

This comes from $RG={1\over3}RB$ as $\large{{RG\over GB}= {RF\over AB} = {1\over2}}$

$\vec{OG}={1\over2}(c+a)+{1\over3}(b-{1\over2}(c+a))={1\over3}(a+b+c)$