If G be the centroid of a triangle ABC and O be any other point, prove that $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=B{{C}^{2}}+C{{A}^{2}}+A{{B}^{2}}$ and $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$?
Answer
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Hint: We start solving the problem by assigning the coordinates to the points for the vertices of triangle ABC. We then find the coordinates of the centroid of the triangle using the fact that the centroid G of a triangle is defined as $\dfrac{A+B+C}{3}$. We then find the required distances using the distance between two points $\left( a,b \right)$ and $\left( c,d \right)$ is defined as $\sqrt{{{\left( a-c \right)}^{2}}+{{\left( b-d \right)}^{2}}}$ and comparing the obtained results to complete the proof $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=B{{C}^{2}}+C{{A}^{2}}+A{{B}^{2}}$. Similarly, we assume the coordinates to the point O and follow the similar procedure to prove $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$.
Complete step-by-step solution
According to the problem, G be the centroid of a triangle ABC and O be any other point. We need to prove $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=B{{C}^{2}}+C{{A}^{2}}+A{{B}^{2}}$ and $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$.
Let us assume the vertices A, B and C be $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$.
We know that the centroid G of a triangle is defined as $\dfrac{A+B+C}{3}$.
So, we get $G=\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$.
We know that the distance between two points $\left( a,b \right)$ and $\left( c,d \right)$ is defined as $\sqrt{{{\left( a-c \right)}^{2}}+{{\left( b-d \right)}^{2}}}$.
So, we get $A{{B}^{2}}={{\left( \sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \right)}^{2}}$.
$\Rightarrow A{{B}^{2}}=x_{1}^{2}+x_{2}^{2}-2{{x}_{1}}{{x}_{2}}+y_{1}^{2}+y_{2}^{2}-2{{y}_{1}}{{y}_{2}}$.
$\Rightarrow A{{B}^{2}}=\left( x_{1}^{2}+x_{2}^{2} \right)+\left( y_{1}^{2}+y_{2}^{2} \right)-2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)$.
Similarly, we get $B{{C}^{2}}=\left( x_{2}^{2}+x_{3}^{2} \right)+\left( y_{2}^{2}+y_{3}^{2} \right)-2\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)$ and $C{{A}^{2}}=\left( x_{3}^{2}+x_{1}^{2} \right)+\left( y_{3}^{2}+y_{1}^{2} \right)-2\left( {{x}_{3}}{{x}_{1}}+{{y}_{3}}{{y}_{1}} \right)$.
Let us consider $A{{B}^{2}}+B{{C}^{2}}+C{{A}^{2}}$.
\[\Rightarrow A{{B}^{2}}+B{{C}^{2}}+C{{A}^{2}}=2\left( x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \right)+2\left( y_{1}^{2}+y_{2}^{2}+y_{3}^{2} \right)-2\left( {{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{1}}+{{y}_{1}}{{y}_{2}}+{{y}_{2}}{{y}_{3}}+{{y}_{3}}{{y}_{1}} \right)\]---(1).
Now, let us find $G{{A}^{2}}$, $G{{B}^{2}}$ and $G{{C}^{2}}$.
So, we get \[G{{A}^{2}}={{\left( \sqrt{{{\left( {{x}_{1}}-\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right) \right)}^{2}}+{{\left( {{y}_{1}}-\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right) \right)}^{2}}} \right)}^{2}}\].
\[\Rightarrow G{{A}^{2}}={{\left( \dfrac{2{{x}_{1}}-{{x}_{2}}-{{x}_{3}}}{3} \right)}^{2}}+{{\left( \dfrac{2{{y}_{1}}-{{y}_{2}}-{{y}_{3}}}{3} \right)}^{2}}\].
\[\Rightarrow G{{A}^{2}}=\left( \dfrac{4x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-4{{x}_{1}}{{x}_{2}}+2{{x}_{2}}{{x}_{3}}-4{{x}_{1}}{{x}_{3}}}{9} \right)+\left(\dfrac{4y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-4{{y}_{1}}{{y}_{2}}-2{{y}_{2}}{{y}_{3}}-4{{y}_{1}}{{y}_{3}}}{9} \right)\].
\[\Rightarrow G{{A}^{2}}=\left( \dfrac{4\left( x_{1}^{2}+y_{1}^{2} \right)+\left( x_{2}^{2}+y_{2}^{2} \right)+\left( x_{3}^{2}+y_{3}^{2} \right)-4\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)+2\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)-4\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{9} \right)\].
Similarly, we get \[G{{B}^{2}}=\left( \dfrac{\left( x_{1}^{2}+y_{1}^{2} \right)+4\left( x_{2}^{2}+y_{2}^{2} \right)+\left( x_{3}^{2}+y_{3}^{2} \right)-4\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-4\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)+2\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{9} \right)\],
\[\Rightarrow G{{C}^{2}}=\left( \dfrac{\left( x_{1}^{2}+y_{1}^{2} \right)+\left( x_{2}^{2}+y_{2}^{2} \right)+4\left( x_{3}^{2}+y_{3}^{2} \right)+2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-4\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)-4\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{9} \right)\].
Now, let us consider $G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}$.
$G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}=\left( \dfrac{6\left( x_{1}^{2}+y_{1}^{2} \right)+6\left( x_{2}^{2}+y_{2}^{2} \right)+6\left( x_{3}^{2}+y_{3}^{2} \right)-6\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-6\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)-6\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{9} \right)$.
$\Rightarrow G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}=\left( \dfrac{2\left( x_{1}^{2}+y_{1}^{2} \right)+2\left( x_{2}^{2}+y_{2}^{2} \right)+2\left( x_{3}^{2}+y_{3}^{2} \right)-2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-2\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)-2\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{3} \right)$---(2).
$\Rightarrow 3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=2\left( x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2} \right)-2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+{{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}}+{{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)$.
From equation (1), we get $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=A{{B}^{2}}+B{{C}^{2}}+A{{C}^{2}}$.
So, we have proved $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=A{{B}^{2}}+B{{C}^{2}}+A{{C}^{2}}$ ---(3).
Now, let us assume the point O be $\left( x,y \right)$.
Now, let us find $O{{A}^{2}}$.
So, we get $O{{A}^{2}}={{\left( \sqrt{{{\left( {{x}_{1}}-x \right)}^{2}}+{{\left( {{y}_{1}}-y \right)}^{2}}} \right)}^{2}}$.
$\Rightarrow O{{A}^{2}}=x_{1}^{2}+{{x}^{2}}-2x{{x}_{1}}+y_{1}^{2}+{{y}^{2}}-2y{{y}_{1}}$.
$\Rightarrow O{{A}^{2}}=\left( x_{1}^{2}+y_{1}^{2} \right)+\left( {{x}^{2}}+{{y}^{2}} \right)-2\left( x{{x}_{1}}+y{{y}_{1}} \right)$.
Similarly, we get $O{{B}^{2}}=\left( x_{2}^{2}+y_{2}^{2} \right)+\left( {{x}^{2}}+{{y}^{2}} \right)-2\left( x{{x}_{2}}+y{{y}_{2}} \right)$ and $O{{A}^{2}}=\left( x_{3}^{2}+y_{3}^{2} \right)+\left( {{x}^{2}}+{{y}^{2}} \right)-2\left( x{{x}_{3}}+y{{y}_{3}} \right)$.
Let us consider $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}$.
\[\Rightarrow O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=\left( x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2} \right)+3\left( {{x}^{2}}+{{y}^{2}} \right)-2\left( x{{x}_{1}}+x{{x}_{2}}+x{{x}_{3}}+y{{y}_{1}}+y{{y}_{2}}+y{{y}_{3}} \right)\].
\[\Rightarrow O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=\left( x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2} \right)+3\left( {{x}^{2}}+{{y}^{2}} \right)-2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)-2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)\]---(4).
Now, let us find $G{{O}^{2}}$.
\[\Rightarrow G{{O}^{2}}={{\left( \sqrt{{{\left( x-\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right) \right)}^{2}}+{{\left( y-\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right) \right)}^{2}}} \right)}^{2}}\].
\[\Rightarrow G{{O}^{2}}={{x}^{2}}+{{\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right)}^{2}}-2x\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right)+{{y}^{2}}+{{\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)}^{2}}-2y\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)\].
\[\Rightarrow G{{O}^{2}}={{x}^{2}}+{{y}^{2}}+\left( \dfrac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+2{{x}_{1}}{{x}_{2}}+2{{x}_{2}}{{x}_{3}}+2{{x}_{1}}{{x}_{3}}+2{{y}_{1}}{{y}_{2}}+2{{y}_{2}}{{y}_{3}}+2{{y}_{1}}{{y}_{3}}}{9} \right)-\left( \dfrac{2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)+2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)}{3} \right)\].
\[\Rightarrow 3G{{O}^{2}}=3\left( {{x}^{2}}+{{y}^{2}} \right)+\left( \dfrac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+2{{x}_{1}}{{x}_{2}}+2{{x}_{2}}{{x}_{3}}+2{{x}_{1}}{{x}_{3}}+2{{y}_{1}}{{y}_{2}}+2{{y}_{2}}{{y}_{3}}+2{{y}_{1}}{{y}_{3}}}{3} \right)-2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)-2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)\].
Using equation (2), we get
$G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}=3\left( {{x}^{2}}+{{y}^{2}} \right)+\left( \dfrac{3\left( x_{1}^{2}+y_{1}^{2} \right)+3\left( x_{2}^{2}+y_{2}^{2} \right)+3\left( x_{3}^{2}+y_{3}^{2} \right)}{3} \right)-2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)-2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)$.
$\Rightarrow G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}=3\left( {{x}^{2}}+{{y}^{2}} \right)+x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2}-2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)-2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)$.
From equation (4), we get $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$ ---(5).
From equations (3) and (5), we have proved $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=B{{C}^{2}}+C{{A}^{2}}+A{{B}^{2}}$ and
$\Rightarrow O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$.
Note: We should not make calculation mistakes while solving this problem. We can consider the point O as origin $\left( 0,0 \right)$ while proving the result $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$ to reduce the confusion and for making the calculation easy. We can also use the length of the medians to prove the given results but it requires a huge amount of knowledge about the properties of medians and their lengths.
Complete step-by-step solution
According to the problem, G be the centroid of a triangle ABC and O be any other point. We need to prove $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=B{{C}^{2}}+C{{A}^{2}}+A{{B}^{2}}$ and $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$.
Let us assume the vertices A, B and C be $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$.
We know that the centroid G of a triangle is defined as $\dfrac{A+B+C}{3}$.
So, we get $G=\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$.
We know that the distance between two points $\left( a,b \right)$ and $\left( c,d \right)$ is defined as $\sqrt{{{\left( a-c \right)}^{2}}+{{\left( b-d \right)}^{2}}}$.
So, we get $A{{B}^{2}}={{\left( \sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \right)}^{2}}$.
$\Rightarrow A{{B}^{2}}=x_{1}^{2}+x_{2}^{2}-2{{x}_{1}}{{x}_{2}}+y_{1}^{2}+y_{2}^{2}-2{{y}_{1}}{{y}_{2}}$.
$\Rightarrow A{{B}^{2}}=\left( x_{1}^{2}+x_{2}^{2} \right)+\left( y_{1}^{2}+y_{2}^{2} \right)-2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)$.
Similarly, we get $B{{C}^{2}}=\left( x_{2}^{2}+x_{3}^{2} \right)+\left( y_{2}^{2}+y_{3}^{2} \right)-2\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)$ and $C{{A}^{2}}=\left( x_{3}^{2}+x_{1}^{2} \right)+\left( y_{3}^{2}+y_{1}^{2} \right)-2\left( {{x}_{3}}{{x}_{1}}+{{y}_{3}}{{y}_{1}} \right)$.
Let us consider $A{{B}^{2}}+B{{C}^{2}}+C{{A}^{2}}$.
\[\Rightarrow A{{B}^{2}}+B{{C}^{2}}+C{{A}^{2}}=2\left( x_{1}^{2}+x_{2}^{2}+x_{3}^{2} \right)+2\left( y_{1}^{2}+y_{2}^{2}+y_{3}^{2} \right)-2\left( {{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{1}}+{{y}_{1}}{{y}_{2}}+{{y}_{2}}{{y}_{3}}+{{y}_{3}}{{y}_{1}} \right)\]---(1).
Now, let us find $G{{A}^{2}}$, $G{{B}^{2}}$ and $G{{C}^{2}}$.
So, we get \[G{{A}^{2}}={{\left( \sqrt{{{\left( {{x}_{1}}-\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right) \right)}^{2}}+{{\left( {{y}_{1}}-\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right) \right)}^{2}}} \right)}^{2}}\].
\[\Rightarrow G{{A}^{2}}={{\left( \dfrac{2{{x}_{1}}-{{x}_{2}}-{{x}_{3}}}{3} \right)}^{2}}+{{\left( \dfrac{2{{y}_{1}}-{{y}_{2}}-{{y}_{3}}}{3} \right)}^{2}}\].
\[\Rightarrow G{{A}^{2}}=\left( \dfrac{4x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-4{{x}_{1}}{{x}_{2}}+2{{x}_{2}}{{x}_{3}}-4{{x}_{1}}{{x}_{3}}}{9} \right)+\left(\dfrac{4y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-4{{y}_{1}}{{y}_{2}}-2{{y}_{2}}{{y}_{3}}-4{{y}_{1}}{{y}_{3}}}{9} \right)\].
\[\Rightarrow G{{A}^{2}}=\left( \dfrac{4\left( x_{1}^{2}+y_{1}^{2} \right)+\left( x_{2}^{2}+y_{2}^{2} \right)+\left( x_{3}^{2}+y_{3}^{2} \right)-4\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)+2\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)-4\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{9} \right)\].
Similarly, we get \[G{{B}^{2}}=\left( \dfrac{\left( x_{1}^{2}+y_{1}^{2} \right)+4\left( x_{2}^{2}+y_{2}^{2} \right)+\left( x_{3}^{2}+y_{3}^{2} \right)-4\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-4\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)+2\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{9} \right)\],
\[\Rightarrow G{{C}^{2}}=\left( \dfrac{\left( x_{1}^{2}+y_{1}^{2} \right)+\left( x_{2}^{2}+y_{2}^{2} \right)+4\left( x_{3}^{2}+y_{3}^{2} \right)+2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-4\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)-4\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{9} \right)\].
Now, let us consider $G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}$.
$G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}=\left( \dfrac{6\left( x_{1}^{2}+y_{1}^{2} \right)+6\left( x_{2}^{2}+y_{2}^{2} \right)+6\left( x_{3}^{2}+y_{3}^{2} \right)-6\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-6\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)-6\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{9} \right)$.
$\Rightarrow G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}=\left( \dfrac{2\left( x_{1}^{2}+y_{1}^{2} \right)+2\left( x_{2}^{2}+y_{2}^{2} \right)+2\left( x_{3}^{2}+y_{3}^{2} \right)-2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}} \right)-2\left( {{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}} \right)-2\left( {{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)}{3} \right)$---(2).
$\Rightarrow 3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=2\left( x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2} \right)-2\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+{{x}_{2}}{{x}_{3}}+{{y}_{2}}{{y}_{3}}+{{x}_{1}}{{x}_{3}}+{{y}_{1}}{{y}_{3}} \right)$.
From equation (1), we get $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=A{{B}^{2}}+B{{C}^{2}}+A{{C}^{2}}$.
So, we have proved $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=A{{B}^{2}}+B{{C}^{2}}+A{{C}^{2}}$ ---(3).
Now, let us assume the point O be $\left( x,y \right)$.
Now, let us find $O{{A}^{2}}$.
So, we get $O{{A}^{2}}={{\left( \sqrt{{{\left( {{x}_{1}}-x \right)}^{2}}+{{\left( {{y}_{1}}-y \right)}^{2}}} \right)}^{2}}$.
$\Rightarrow O{{A}^{2}}=x_{1}^{2}+{{x}^{2}}-2x{{x}_{1}}+y_{1}^{2}+{{y}^{2}}-2y{{y}_{1}}$.
$\Rightarrow O{{A}^{2}}=\left( x_{1}^{2}+y_{1}^{2} \right)+\left( {{x}^{2}}+{{y}^{2}} \right)-2\left( x{{x}_{1}}+y{{y}_{1}} \right)$.
Similarly, we get $O{{B}^{2}}=\left( x_{2}^{2}+y_{2}^{2} \right)+\left( {{x}^{2}}+{{y}^{2}} \right)-2\left( x{{x}_{2}}+y{{y}_{2}} \right)$ and $O{{A}^{2}}=\left( x_{3}^{2}+y_{3}^{2} \right)+\left( {{x}^{2}}+{{y}^{2}} \right)-2\left( x{{x}_{3}}+y{{y}_{3}} \right)$.
Let us consider $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}$.
\[\Rightarrow O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=\left( x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2} \right)+3\left( {{x}^{2}}+{{y}^{2}} \right)-2\left( x{{x}_{1}}+x{{x}_{2}}+x{{x}_{3}}+y{{y}_{1}}+y{{y}_{2}}+y{{y}_{3}} \right)\].
\[\Rightarrow O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=\left( x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2} \right)+3\left( {{x}^{2}}+{{y}^{2}} \right)-2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)-2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)\]---(4).
Now, let us find $G{{O}^{2}}$.
\[\Rightarrow G{{O}^{2}}={{\left( \sqrt{{{\left( x-\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right) \right)}^{2}}+{{\left( y-\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right) \right)}^{2}}} \right)}^{2}}\].
\[\Rightarrow G{{O}^{2}}={{x}^{2}}+{{\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right)}^{2}}-2x\left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right)+{{y}^{2}}+{{\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)}^{2}}-2y\left( \dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)\].
\[\Rightarrow G{{O}^{2}}={{x}^{2}}+{{y}^{2}}+\left( \dfrac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+2{{x}_{1}}{{x}_{2}}+2{{x}_{2}}{{x}_{3}}+2{{x}_{1}}{{x}_{3}}+2{{y}_{1}}{{y}_{2}}+2{{y}_{2}}{{y}_{3}}+2{{y}_{1}}{{y}_{3}}}{9} \right)-\left( \dfrac{2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)+2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)}{3} \right)\].
\[\Rightarrow 3G{{O}^{2}}=3\left( {{x}^{2}}+{{y}^{2}} \right)+\left( \dfrac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+2{{x}_{1}}{{x}_{2}}+2{{x}_{2}}{{x}_{3}}+2{{x}_{1}}{{x}_{3}}+2{{y}_{1}}{{y}_{2}}+2{{y}_{2}}{{y}_{3}}+2{{y}_{1}}{{y}_{3}}}{3} \right)-2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)-2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)\].
Using equation (2), we get
$G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}=3\left( {{x}^{2}}+{{y}^{2}} \right)+\left( \dfrac{3\left( x_{1}^{2}+y_{1}^{2} \right)+3\left( x_{2}^{2}+y_{2}^{2} \right)+3\left( x_{3}^{2}+y_{3}^{2} \right)}{3} \right)-2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)-2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)$.
$\Rightarrow G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}=3\left( {{x}^{2}}+{{y}^{2}} \right)+x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+x_{3}^{2}+y_{3}^{2}-2x\left( {{x}_{1}}+{{x}_{2}}+{{x}_{3}} \right)-2y\left( {{y}_{1}}+{{y}_{2}}+{{y}_{3}} \right)$.
From equation (4), we get $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$ ---(5).
From equations (3) and (5), we have proved $3\left( G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}} \right)=B{{C}^{2}}+C{{A}^{2}}+A{{B}^{2}}$ and
$\Rightarrow O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$.
Note: We should not make calculation mistakes while solving this problem. We can consider the point O as origin $\left( 0,0 \right)$ while proving the result $O{{A}^{2}}+O{{B}^{2}}+O{{C}^{2}}=G{{A}^{2}}+G{{B}^{2}}+G{{C}^{2}}+3G{{O}^{2}}$ to reduce the confusion and for making the calculation easy. We can also use the length of the medians to prove the given results but it requires a huge amount of knowledge about the properties of medians and their lengths.
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