Question

prove that the sum of the squares of the sides of an rhombus is equal to the sum of the squares of its diagonals

Answer 1

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)

⇒ AB2 + BC2 + CD2 + AD2 = AC2 + BD2

Answer 2

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)

⇒ AB2 + BC2 + CD2 + AD2 = AC2 + BD2