Question

How do you prove that sum of the squares on the sides of a rhombus is equal to the sum of squares on its diagonals?

Answer 1

see explanation.

Explanation:

Some of the properties of a rhombus :
`a)` all sides have equal length :
`=> AB=BC=CD=DA`
`b)` the two diagonals bisect each other at right angle:
`=> OA=OC, => AC=2OA=2OC`
Similarly, `OB=OD, => BD=2OB=2OD`

In `DeltaAOB`, as `angleAOB=90^@`,
by Pythagorean theorem, we know that `AB^2=OA^2+OB^2`
Let the sum of squares of the sides be `S_s`,
`=> S_s=4AB^2`

Let the sum of squares of the diagonals be `S_d`,
`S_d` is given by :
`S_d=AC^2+BD^2=(2OA)^2+(2OB)^2`
`=> S_d=4OA^2+4OB^2=4(OA^2+OB^2)=4AB^2`

`=> S_d=S_s` (proved)