lf G be the centroid of a triangle ABC and P be any other point in the plane prove that `PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3GP^2`

Explanation

Let the points be `A(x1​,y1​),B(x2​,y2​),C(x3​,y3​)`

and the point `P` be `(0,0,0)`

Taking L.H.S.

`(PA)^2+(PB)^2+(PC)^2 = (x1)^2​+(x2)^2​+(x3)^2​+(y1)^2​+(y2)^2​+(y3)^2​ ->(1)`


Cordinates of centroid

`G=((x1​+x2​+x3)/3​​,(y1​+y2​+y3​)/3​)
`

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