Proving OG = 1/3(OA + OB + OC) for a triangle centroid

[eekoz]

Okay, so I have a triagle with vertices A, B, and C.
I know that the centroid, G, is where all the medians of the triangle intersect, and G divides the median at a 2:1 ratio

Assuming point O is a point that's not on the triangle, how can I prove:
OG = 1/3(OA + OB + OC) ?

I've seen this equation a lot of times, but I'd like to see a proof of it for interest sake

Thanks

 

[AKG]

Let $P$ be the midpoint of [itex]\overline{BC}[/tex]. Then:<br /> <br /> $$\overline{OB} + \frac{1}{2}\overline{BC} = \overline{OP}$$<br /> <br /> $$\overline{OB} + \overline{BC} = \overline{OC}$$<br /> <br /> $$\overline{OP} + \overline{PA} = \overline{OA}$$<br /> <br /> $$\overline{OP} + \frac{1}{3}\overline{PA} = \overline{OG}$$<br /> <br /> The last line comes from the fact that G divides the median $\overline{PA}$ at a 2:1 ratio. You should be able to figure it out from here.[/itex]