Question

Question 7
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 whose diagonals AC and BD intersect at O.
We have to prove that $A{B}^2+B{C}^2+C{D}^2+A{D}^2=A{C}^2+B{D}^2$
Since the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
In $\Delta AOB$,
$\angle AOB={90}^{\circ }$
$A{B}^2=A{O}^2+B{O}^2...\left(i\right)\left[ByPythagorastheorem\right]$
Similarly,
$A{D}^2=A{O}^2+D{O}^2...\left(ii\right)$
$D{C}^2=D{O}^2+C{O}^2...\left(iii\right)$
$B{C}^2=C{O}^2+B{O}^2...\left(iv\right)$

Adding equations (i) + (ii) + (iii) + (iv) , we get,
$A{B}^2+A{D}^2+D{C}^2+B{C}^2=2(A{O}^2+B{O}^2+D{O}^2+C{O}^2)$
$=4A{O}^2+4B{O}^2$ [Since, AO = CO and BO =DO]
$=(2AO{)}^2+(2BO{)}^2=A{C}^2+B{D}^2$

$Hence,A{B}^2+B{C}^2+C{D}^2+A{D}^2=A{C}^2+B{D}^2$
Hence proved.