Question

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

Answer 1

STEP 1 : Assumption

Let $ABCD$ be a rhombus with diagonals $AC$ and $BD$ intersecting at $O$ as shown in the figure.

We know that sides of a rhombus are equal.

i.e. $AB=BC=CD=DA$

We also know that diagonals of a rhombus bisect each other at right angles.

i.e. $AO=OC,BO=OD$ and $AC\perp BD$

STEP 2 : Proving that $A{B}^2+B{C}^2+C{D}^2+D{A}^2=A{C}^2+B{D}^2$

Since, we know that diagonals of a rhombus bisect each other at right angles, we get

$AO=OC=\frac{1}{2}AC$ and

$BO=OD=\frac{1}{2}BD$ $...\left(1\right)$

Now, In $△AOB$, $\angle AOB=90°$

So by using Pythagoras Theorem in $△AOB$, we get

$A{B}^2=A{O}^2+B{O}^2$

From equation $\left(1\right)$

$A{B}^2={\left(\frac{1}{2}AC\right)}^2+{\left(\frac{1}{2}BD\right)}^2$

$A{B}^2={\frac{AC}{4}}^2+{\frac{BD}{4}}^2=\frac{A{C}^2+B{D}^2}{4}$

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

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

We know that $AB=BC=CD=DA$

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

Hence, it is proved that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.