Question

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

Answer 1

STEP 1 : Construction

Let us draw a parallelogram $ABCD$ such that $AB=CD$ and $AD=BC$.

$AC$ and $BD$ are the diagonals of the parallelogram $ABCD$ as shown in figure.

Draw $AE\perp CD$ and $DF\perp AB$

$EA=DF$ (Perpendiculars drawn between same parallel lines)

STEP 2 : Finding the value of $A{C}^2$

In $△AEC$, according to Pythagoras Theorem

$A{C}^2=A{E}^2+E{C}^2$

$A{C}^2=A{E}^2+{\left(ED+DC\right)}^2$ [Since $EC=ED+DC$]

$A{C}^2=A{E}^2+E{D}^2+D{C}^2+2.ED.DC$ $[{\left(a+b\right)}^2=a^2+b^2+2ab]$

$A{C}^2=\left(A{D}^2-E{D}^2\right)+E{D}^2+D{C}^2+2.ED.DC$ $[In△AED,A{E}^2=A{D}^2-E{D}^2]$

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

STEP 3 : Finding the value of $B{D}^2$

In $△DFB$, according to Pythagoras Theorem

$D{B}^2=F{D}^2+F{B}^2$

$D{B}^2=F{D}^2+{\left(AB-AF\right)}^2$ [Since $FB=AB-AF$]

$D{B}^2=F{D}^2+A{B}^2+A{F}^2-2.AB.AF$ $[{\left(a-b\right)}^2=a^2+b^2-2ab]$

$D{B}^2=\left(A{D}^2-A{F}^2\right)+A{B}^2+A{F}^2-2.AB.AF$ $[In△AFD,F{D}^2=A{D}^2-A{F}^2]$

$D{B}^2=A{D}^2+A{B}^2-2.AB.AF$ $...\left(2\right)$

STEP 4 : Finding the value of $A{C}^2+B{D}^2$

Adding equation $\left(1\right)$ and $\left(2\right)$

$A{C}^2+B{D}^2=A{D}^2+D{C}^2+2.ED.DC+A{D}^2+A{B}^2-2.AB.AF$

$A{C}^2+B{D}^2=B{C}^2+D{C}^2+2.AB.AF+A{D}^2+A{B}^2-2.AB.AF$ $[SinceAD=BCandED=AF,DC=AB]$

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

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

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