Question

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer 1

Given:

$ABCD$ is a rhombus.

Hence, $AB=BC=CD=AD$

And, AC perpendicular to BD

$DO=\frac{1}{2}DB$ and $AO=\frac{1}{2}AC$

To Prove:

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

Proof:

In $△AOD$, By Pythagoras Theorem,

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

Similarly,

$D{C}^2=D{O}^2+O{C}^2.....\left(2\right)$

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

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

Adding 1, 2, 3, 4 we get,

$A{B}^2+B{C}^2+C{D}^2+A{D}^2=2A{O}^2+2B{O}^2+2C{O}^2+2D{O}^2....\left(5\right)$

Since, $DO=OB=\frac{1}{2}DB$ and $AO=OC=\frac{1}{2}AC....\left(6\right)$

From 5 and 6,

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

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

Hence Proved.