Question

If $G$ be the centroid of a triangle $ABC$, prove that $A{B}^2+B{C}^2+A{C}^2=3\left(G{A}^2+G{B}^2+G{C}^2\right)$.

Answer 1

Step 1: Find the centroid of the triangle

Given:

• $G$ is the centroid of a triangle $ABC$
• Let $G=(0,0)$ and $A=(x_1,y_1),B=(x_2,y_2),C=(x_3,y_3)$

Centroid of $△ABC=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

$(0,0)=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

Step 2: Find the sides $AB,BC,CA,GA,GB,GC$

so, $x_1+x_2+x_3=0,y_1+y_2+y_3=0$

squaring both sides

$$\begin{gathered} {\left(x_1+x_2+x_3\right)}^2=0,{\left(y_1+y_2+y_3\right)}^2=0 \\ {x}_1^2+x_2^2+x_3^2+2x_1{x}_2+2x_2{x}_3+2x_1{x}_3=0-----\left(1\right) \end{gathered}$$

and $y_1^2+y_2^2+y_3^2+2y_1{y}_2+2y_2{y}_3+2y_1{y}_3=0------\left(2\right)$

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2},BC=\sqrt{{(x_3-x_2)}^2+{(y_3-y_2)}^2},AC=\sqrt{{(x_3-x_1)}^2+{(y_3-y_1)}^2}$

and $$\begin{gathered} GA=\sqrt{{(x_1-0)}^2+{(y_1-0)}^2},GB=\sqrt{{(x_2-0)}^2+{(y_2-0)}^2},GC=\sqrt{{(x_3-0)}^2+{(y_3-0)}^2} \\ \Rightarrow GA=\sqrt{x_1^2+{y_1}^2},GB=\sqrt{x_2^2+{y_2}^2},GC=\sqrt{x_3^2+{y_3}^2} \\ G{A}^2+G{B}^2+G{C}^2=x_1^2+x_2^2+x_3^2+{y_1}^2+{y_2}^2+{y_3}^2------\left(3\right) \end{gathered}$$

Step 3: Simplify the LHS

$$\begin{gathered} A{B}^2+B{C}^2+C{A}^2={\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(x_3-x_2\right)}^2+{\left(y_2-y_1\right)}^2+{(x_3-x_1)}^2+{(y_3-y_1)}^2 \\ A{B}^2+B{C}^2+C{A}^2=x_1^2+x_2^2+x_3^2+x_1^2+x_2^2+x_3^2-2x_1{x}_2-2x_2{x}_3-2x_1{x}_3+y_1^2+y_2^2+y_3^2+y_1^2+y_2^2+y_3^2-2y_1{y}_2-2y_2{y}_3-2y_1{y}_3 \\ A{B}^2+B{C}^2+C{A}^2=2\left(x_1^2+x_2^2+x_3^2\right)+2\left(y_1^2+y_2^2+y_3^2\right)-\left(2x_1{x}_2+2x_2{x}_3+2x_1{x}_3\right)-\left(2y_1{y}_2+2y_2{y}_3+2y_1{y}_3\right) \\ A{B}^2+B{C}^2+C{A}^2=2\left(x_1^2+x_2^2+x_3^2\right)+2\left(y_1^2+y_2^2+y_3^2\right)+\left(x_1^2+x_2^2+x_3^2\right)+\left(y_1^2+y_2^2+y_3^2\right)\left[u\sin gequation\left(1\right)and\left(2\right)\right] \\ A{B}^2+B{C}^2+C{A}^2=3\left(x_1^2+x_2^2+x_3^2\right)+3\left(y_1^2+y_2^2+y_3^2\right) \\ A{B}^2+B{C}^2+C{A}^2=3\left(x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2\right) \\ A{B}^2+B{C}^2+C{A}^2=3\left(G{A}^2+G{B}^2+G{C}^2\right)\left[u\sin gequation\left(3\right)\right] \end{gathered}$$

Hence, $A{B}^2+B{C}^2+A{C}^2=3\left(G{A}^2+G{B}^2+G{C}^2\right)$ is proved.