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

Explanation

Let ABCD be a rhombus whose eachd side is a. the diagonals of rhombus bisects perpendicularly at O.
Now, in right `triangleAOB`
`OA^(2) + OB^(2)=a^(2) ` ( by pythagores theorem)
` (1/2 AC)^(2) + (1/2BD)^(2) =a ^(2)` ( diagonals bisect each other)
` 1/4AC^(2)+1/4BD^(2)=a^(2)`
` Rightarrow " " AC^(2) + BD^(2)= 4a^(2)`
` Rightarrow AC^(2) + BD^(2)= a^(2)+a^(2)+a^(2)+a^(2)`
` Rightarrow " " AC^(2) + BD^(2) = AB^(2)+ BC^(2) +CD ^(2) + DA^(2)` Hence proved.