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

Let $\square ABCD$ be a parallelogram.

Let its diagonals $AC$ and $BD$ intersect at O.

In $△ABC$,

BO is the median ....Diagonals of a parallelogram bisect each other

$\therefore$ By Apollonius theorem,

$A{B}^2+B{C}^2=2O{B}^2+2O{A}^2$ ....(1)

In $△ADC$,

DO is the median ....Since diagonals bisect each other

$\therefore$ By Apollonius theorem,

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

Adding (1) and (2) we get,

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

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

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

$A{B}^2+B{C}^2+C{D}^2+A{D}^2=4{(\frac{1}{2}\times DB)}^2+4{(\frac{1}{2}\times CA)}^2$ [$\because OB=\frac{DB}{2}andOA=\frac{CA}{2}$]

$A{B}^2+B{C}^2+C{D}^2+A{D}^2=4(\frac{1}{4}\times D{B}^2)+4(\frac{1}{4}\times C{A}^2)$

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