Question

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

Answer 1

REF.Image

$AD$ is the median

In triangle AXB & AXC, by pythagoras theorem,

$A{B}^2=A{X}^2+B{X}^2$ (1)

and $A{C}^2=A{X}^2+C{X}^2$ (2)

Adding (1) & (2)

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

but $BX=BD-DX$ and $CX=DX+DC=BD+DX$

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

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

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

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

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

$A{B}^2+A{C}^2=2(A{X}^2+D{X}^2)+2B{D}^2$ $\because A{X}^2+D{X}^2=A{D}^2$

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

$A{B}^2+A{C}^2=2A{D}^2+\frac{B{C}^2}{2}$ (3)

similarly, $B{C}^2+A{B}^2=2B{E}^2+\frac{A{C}^2}{2}$ (4)

$A{C}^2+B{C}^2=2C{F}^2+\frac{A{B}^2}{2}$ (5)

Adding (3),(4) and (5)

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

multiplying throughout by 2,

$4(A{B}^2+B{C}^2+A{C}^2)=4(A{D}^2+B{E}^2+C{F}^2)+(A{B}^2+B{C}^2+A{C}^2)$

$3(A{B}^2+B{C}^2+A{C}^2)=4(A{D}^2+B{E}^2+C{F}^2)$

hence Proved.