Prove that the sum of squares of a sides of a rhombus is equal to the sum of the squares of its diagonals......pls give answer.....u'll surely get a ''thumbs up''
Consider a rhombus ABCD, whose diagonals intersect at O.
We know that diagonals of a rhombus bisect each other at 90°, so triangles AOB, AOD, BOC and COD are right angled triangles.
In ΔAOB,
AO2 + OB2 = AB2 (Using Pythagoras theorem)
⇒ AC2 + BD2 = 4AB2
Similarly, AC2 + BD2 = 4BC2
AC2 + BD2 = 4CD2
AC2 + BD2 = 4AD2
Adding all these equations, we get
4(AB2 + BC2 + CD2 + AD2) = 4(AC2 + BD2)
Hence AB2 + BC2 + CD2 + AD2 = AC2 + BD2