Question

Prove that the sum of the squares on the three sides of an equilateral triangle is equal to four times the square on a median.

Answer 1

R.E.F image

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

$=2\times \frac{1}{2}B{C}^2+2A{D}^2$

$=\frac{1}{2}B{C}^2+2A{D}^2$

$2A{B}^2+2A{C}^2=B{C}^2+4A{D}^2...\left(1\right)$

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

$2B{C}^2+2A{C}^2=A{B}^2+4C{F}^2...\left(3\right)$

$\left(1\right)+\left(2\right)+\left(3\right)$

$4A{B}^2+4B{C}^2+4A{C}^2=A{B}^2+B{C}^2+A{C}^2+4A{D}^2$

$+4B{E}^2+4C{F}^2$

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