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If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA² +PB²+ PC² = GA² + GB²+ GC² + 3GP².
You can do it by coordinates and vectors both methods vector method is simple and involves less variables assume P as origin Position vectors of A,B,C as a,b,c respectively PA=a, PB=b,OPC=c G=a+b+c/3 GA=a-(a+b+c)/3=2a-b-c/3 similarly you can write for GB,GC,GP GA2=(4a2+b2+c2-4a.b+b.c-4a.c)/9 similarly write for others and just calculate the expression similarly using coordinate you can take P as (0,0)and A,B,C as (x1,y1),(x2,y2)and (x3,y3) G will be ((x1+x2+x3)/3 , (y1+y2+y3)/3) and using distance formula solve the expressions on LHS and RHS
You can do it by coordinates and vectors both methods
vector method is simple and involves less variables
assume P as origin
Position vectors of A,B,C as a,b,c respectively
PA=a, PB=b,OPC=c
G=a+b+c/3
GA=a-(a+b+c)/3=2a-b-c/3
similarly you can write for GB,GC,GP
GA2=(4a2+b2+c2-4a.b+b.c-4a.c)/9
similarly write for others and just calculate the expression
similarly using coordinate
you can take P as (0,0)and A,B,C as (x1,y1),(x2,y2)and (x3,y3)
G will be ((x1+x2+x3)/3 , (y1+y2+y3)/3)
and using distance formula solve the expressions on LHS and RHS
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