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

(N/A) Let $ABCD$ be a rhombus with diagonals $AC$ and $BD$ intersecting at $O$.
In $\Delta AOB, \Delta BOC, \Delta COD,$ and $\Delta AOD,$
Applying the Pythagoras theorem,we obtain:
$AB^2 = AO^2 + OB^2$ $...(1)$
$BC^2 = BO^2 + OC^2$ $...(2)$
$CD^2 = CO^2 + OD^2$ $...(3)$
$AD^2 = AO^2 + OD^2$ $...(4)$
Adding all these equations,we obtain:
$AB^2 + BC^2 + CD^2 + AD^2 = 2(AO^2 + OB^2 + OC^2 + OD^2)$
Since the diagonals of a rhombus bisect each other at right angles,$AO = OC = AC/2$ and $BO = OD = BD/2$.
Substituting these values:
$= 2((AC/2)^2 + (BD/2)^2 + (AC/2)^2 + (BD/2)^2)$
$= 2(2(AC/2)^2 + 2(BD/2)^2)$
$= 2(AC^2/2 + BD^2/2)$
$= AC^2 + BD^2$
Thus,the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.