Question

Prove that the sum of the squares of the sides of a Rhombus is equal to the sum of the squares ondiagonals.​

Answer 1

$a$ - the side

$d_1,d_2$ - the diagonals

By the Pythagorean theorem

$a^2=\left(\dfrac{1}{2}d_1\right)^2+\left(\dfrac{1}{2}d_2\right)^2\\\\a^2=\dfrac{d_1^2}{4}+\dfrac{d_2^2}{4}\\\\\boxed{4a^2=d_1^2+d_2^2}$