If G is the centroid of a `DeltaABC`, then `GA^(2)+GB^(2)+GC^(2)` is equal to

Options

A. `a^(2)+b^(2)+c^(2)`
B. `(a^(2)+b^(2)+c^(2))/3`
C. `(a^(2)+b^(2)+c^(2))/2`
D. `((a+b+c)^(2))/3`

Explanation

In Fig. 30, we have
`GB^(2) + GC^(2) = 2 (GD^(2) + BD^(2))`
`rArrGB^(2)+GC^(2)=2(1/4GA^(2)+1/4a^(2))`
`rArrGB^(2)+GC^(2)=1/2GA^(2)+(a^(2))/2`
Similarly, we have
`GA^(2)+GB^(2)=1/2GC^(2)+(c^(2))/2` and `GC^(2)+GA^(2)=1/2GB^(2)+(b^(2))/2`
Adding these results, we get
`2(GA^(2)+GB^(2)+GC^(2))=1/2(GA^(2)+GB^(2)+GC^(2))+1/2(a^(2)+b^(2)+c^(2))`
`rArrGA^(2)+GB^(2)+GC^(2)=(a^(2)+b^(2)+c^(2))/3`