given the point g is the centre of gravity of triangle ABC,prove the following identity?

-----------A


-----P--------------R
-------------G

B------------Q--------------C
P is an intersection of AB and CG, AP=PB
Q is an intersection of BC and AG, BQ=QC
R is an intersection of AC and BG, CR=RA

From midpoint theorem, BC//PR.
So BC:PR=BG:GR=CG:GP=2:1. Similarly, AG:GQ=2:1.
That is, AQ=(3/2)AG, BR=(3/2)BG, CP=(3/2)CG.

From parallelogram law and avobe result,
AB^2+CA^2=2(AQ^2+BQ^2)
=(9/2)AG^2+(1/2)BC^2 ---(1)

AB^2+BC^2=2(BR^2+CR^2)
=(9/2)BG^2+(1/2)CA^2 ---(2)

BC^2+CA^2=2(CP^2+AP^2)
=(9/2)CG^2+(1/2)AB^2 ---(3)

From (1)+(2)+(3),
2AB^2+2BC^2+2CA^2
=(9/2)(AG^2+BG^2+CG^2)
+(1/2)(AB^2+BC^2+CA^2)

(3/2)(AB^2+BC^2+CA^2)
=(9/2)(AG^2+BG^2+CG^2)

(1/3)(AB^2+BC^2+CA^2)
=AG^2+BG^2+CG^2 Q.E.D.