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Question

If G be the centroid of a triangle ABC, prove that AB2+BC2+CA2=3(GA2+GB2+GC2).

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Solution

LetA(x1,y1),B(x2,y2)andC(x3,y3)betheverticesofABC.
Withoutthelawofgenerality,assumethecentroidofthe
ABCtobeatorigin,i.e.,G=(0,0).
CentroidofABC=[x1+x2+x33,y1+y2+y33]
x1+x2+x3=0;y1+y2+y3=0
Squaringonbothsides,
x12+x22+x32+2x1.x2+2x2.x3+2x3.x1=0
y12+y22+y32+2y1.y2+2y2.y3+2y3.y1=0(i)
AB2+BC2+CA2
=[(x2x1)2+(y2y1)2]+[(x3x2)2+(y3y2)2]+[(x1x3)2+(y1y3)2]
=(x12+x222x1x2+y12+y222y1y2)+(x12+x322x1x3+y12+y322y2y3)
+(x12+x322x1x3+y12+y322y1y3)
=(2x12+2x22+2x322x1x22x2x32x1x3)+
(2y12+2y22+2y32y1y22y2y32y1y3)
=(3x12+3x22+3x32)+(3y12+3y22+3y32)
=3(x12+x22+x32)+3(y12+y22+y32)(ii)
3(GA2+GB2+GC2)
=3[(x10)2+(y10)2+(x20)2+(y20)2+(x30)2+(y30)2]
=3[x12+y12+x22+y22+x32+y32]
=3(x12+x22+x32)+3(y12+y22+y32)(iii)
from(ii)&(iii)
AB2+BC2+CA2=3(GA2+GB2+GC2)

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