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Question
If G be the centroid of a triangle ABC, prove that:
AB2 + BC2 + CA2 = 3 (GA2 + GB2 + GC2)
Solution
Let A(x1,y1); B(x2,y2); C(x3,y3) be the coordinates of the vertices of ΔABC.Let us assume that centroid of the ΔABC is at the origin G.So, the coordinates of G are G(0,0).
Now,`(x_1+x_2+x_3)/3 =0; (y_1+y_2+y_3)/3 =0`
So, `x_1+x_2+x_3=0` ...........(1)
`y_1+y_2+y_3=0` ..........(2)
Squaring (1) and (2), we get
`x_1^2+x_2^2+x_3^2+2x_1x_2+2x_2x_3+2x_3x_1=0` ..........(3)
`y_1^2+y_2^2+y_3^2+2y_1y_2+2y_2y_3+2y_3y_1=0 ` ..........(4)
`LHS=AB^2+BC^2+CA^2`
`=[sqrt((x_2-x_1)^2 +(y_2-y_1)^2]]^2 +[sqrt((x_3-x_2)^2+(y_3-y_2)^2)]^2 +[sqrt((x_3-x_1)^2+(y_3-y_1)^2)]^2 `
`=(x_2-x_1)^2 +(y_2-y_1)^2+(x_3-x_2)^2+(y_3-y_2)^2+(x_3-x_1)^2+(y_3-y_1)^2`
`=x_1^2x_2^2-2x_1^2+y_1^2+y_2^2-2y_1y_2+x_2^2+x_3^2-2x_2x_3+y_2^2+y_2^2+y_3^2-2y_2y_3+x_1^2+x_3^2-2x_1x_3+y_1^2+y_3^2-2y_1v_3`
`=2(x_1^2+x_2^2+x_3^2)+2(y_1^2+y_2^2+y_3^2)-(2x_1x_2+2x_2x_3+2x_3x_1)-(2y_1y_2+2y_2y_3+2y_3y_1)`
`=2(x_1^2+x_2^2+x_3^2)+2(y_1^2+y_2^2+y_3^2)+(x_1^2+x_2^2+x_3^2)+(y_1^2+y_2^2+y_3^2)`
`=3(x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2)`
`RHS =3(GA^2+GB^2+GC^2)`
`=[{sqrt((x_1-0)^2+(y_1-0)^2)}^2 +{sqrt((x_2-0)^2+(y_2-0)^2)}^2 +{sqrt((x_3-0)^2+(y_3-0)^2)}^2]`
`=3[x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2]`
Hence, `AB^2+BC^2+CA^2=3(GA^2+GB^2+GC^2)`
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