# If G is the CENTROID of triangle ABC then prove that AB?2 +BC?2+AC?2=3(GA?2+GB?2+GC?2) BEST OF LUCK FOR THE EQUATION

Answers (1)

$Let A(x_1,y_1),B(x_2,y_2)\;and\;C(x_3,y_3),\;be\;the\;vertices\;of\;\triangle ABC\\*Assume\;the\;centroid\;of\;the\; \triangle ABC\;to\;be\;at\;the\;origin\\* Centroid=(0,0)= (\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})\\* \Rightarrow x_1+x_2+ x_3=0\;and\;y_1+y_2+y_3=0\\* squaring\;both\;side,\;we\;get\\*x_1^2+ x_2^2+x_3^2+2x_1x_2+2x_2x_3+2x_3x_1=0,\;and\\*y_1^2+y_2^2+y_3^2+2y_1y_2+2y_2y_3+2y_3y_1=0\;\;\;\;\;\;\;..eq(1)\\*\Rightarrow AB^2+BC^2 +CA^2 \\* =[(x_2-x_1)^2+(y_2-y_1)^2]+[(x_3-x_2)^2+(y_3- y_2)^2]+[(x_1-x_3)^2+(y_1-y_3)^2]\\* = [(x_1^2+x_2^2-2x_1x_2+y_1^2+ y_2^2-2y_1 y_2)+(x_2^2+x_3^2-2x_2x_3+y_2^2+y_3^2-2y_2y_3)+(x_1^2+x_3^2-2 x_1x_3+y_1^2+y_3^2-2y_1y_3)\\*using\;..eq(1),\\*=3(x_1^2+x_2^2+ x_ 3^ 2)+3(y_1^2+ y_2^2+y_3^2)\;\;\;\;\;..eq(2)\\* \Rightarrow 3(GA^2 + GB^2 + GC^2)\\*=3[(x_1-0)^2+(y_1-0)^2+(x_2-0)^2 +(y_2-0)^2+(x_3-0)^2+(y_3-0)^2]\\* =3(x_1^2+x_2^2+x_3^2)+3(y_1^2+ y_2^2+y_3^2)\;\;\;\;..eq(3)\\* from..eq(1)\;and\;..eq(2)\\* AB^2+BC^2+CA^2 =3(GA^2+GB^2+GC^2)$

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