Question

Prove that three times the sum of the square of the sides of a triangle is equal to four times the sum of the square of the medians of the triangle.

Answer 1

From the diagram

$BC$ and $AD$ is the median.

In $△AXBandAXC$ By Pythagoras Theorem,

$A{B}^2=A{X}^2+B{X}^2--\left(i\right)$

And

$A{C}^2=A{X}^2+C{X}^2--\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$ we get

$A{B}^2+A{C}^2=2A{X}^2+B{X}^2+C{X}^2$

But, $BX=BD-DX$ and $CX=DX+DC$

$DX+BD$ since $BD=DC$

$\therefore B{X}^2+C{X}^2={\left(BD-DX\right)}^2+{\left(DX+BD\right)}^2$

$\Rightarrow B{X}^2+C{X}^2=\left(B{D}^2+D{X}^2-2BD\times DX\right)+\left(D{X}^2+B{D}^2+2BD\times DX\right)$

$\Rightarrow B{X}^2+C{X}^2=2B{D}^2+2D{X}^2$

Now

$A{B}^2+A{C}^2=2A{X}^2+B{X}^2+C{X}^2$

$\Rightarrow A{B}^2+A{C}^2=2A{X}^2+2B{D}^2+2D{X}^2$

$\Rightarrow A{B}^2+A{C}^2=2\left(A{X}^2+D{X}^2\right)+2B{D}^2\left\{\therefore A{X}^2+D{X}^2=A{D}^2\right\}$

$\therefore A{B}^2+A{C}^2=2A{D}^2+2B{D}^2$

Now,

$A{B}^2+A{C}^2=2A{D}^2+\frac{B{C}^2}{2}----\left(iii\right)$

Similarly

$B{C}^2+A{B}^2=2B{E}^2+\frac{A{C}^2}{2}-----\left(iv\right)$

$A{C}^2+B{C}^2=2C{F}^2+\frac{A{B}^2}{2}-----\left(v\right)$

Now adding $\left(iii\right),\left(iv\right)and\left(v\right)$ we get

$2\left(A{B}^2+B{C}^2+A{C}^2\right)=2\left(A{D}^2+B{C}^2+C{E}^2\right)+\frac{1}{2}\left(A{B}^2+B{C}^2+A{C}^2\right)$

On multiplying through $2$ we get

$4\left(A{B}^2+B{C}^2+A{C}^2\right)=4\left(A{D}^2+B{C}^2+C{E}^2\right)+\left(A{B}^2+B{C}^2+A{C}^2\right)$

$\Rightarrow 3\left(A{B}^2+B{C}^2+A{C}^2\right)=4\left(A{D}^2+B{C}^2+C{E}^2\right)$

Proved.