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

$ABCD$ is a rhombus with ${}^{\prime }{O}^{\prime }$ as the point of intersection of diagonals.

We know that the diagonals of a rhombus bisect each other at ${90}^{\circ }$.

i.e. $AC\perp BD,OA=OC,OB=OD,AC=\frac{1}{2}OA,BD=\frac{1}{2}OB$

In triangle $OAB,OBC,OCD$ and $OAD$ we have

$A{B}^2=O{A}^2+O{B}^2\dots \dots \left(i\right)$ [Using Pythagoras theorem]

$B{C}^2=O{B}^2+O{C}^2\dots \dots \left(ii\right)$

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

$A{D}^2=O{A}^2+O{D}^2\dots \dots \left(iv\right)$

On adding these these equations, we get

$A{B}^2+B{C}^2+C{D}^2+A{D}^2=2(O{A}^2+O{B}^2+O{C}^2+O{D}^2)$

$A{B}^2+B{C}^2+C{D}^2+A{D}^2=2\left[\right\{\frac{AC}{2}{\}}^2+\{\frac{BD}{2}{\}}^2+\{\frac{AC}{2}{\}}^2+\left\{\frac{BD}{2}{\}}^2\right]$

$A{B}^2+B{C}^2+C{D}^2+A{D}^2=2[\frac{A{C}^2}{4}+\frac{B{D}^2}{4}+\frac{A{C}^2}{4}+\frac{B{D}^2}{4}]$

$A{B}^2+B{C}^2+C{D}^2+A{D}^2=2[\frac{2A{C}^2}{4}+\frac{2B{D}^2}{4}]$

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

Hence, the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.