Question

ABCD is a rhombus. Prove that AB² + BC² + CD² + DA²= AC² + BD²

Answer 1

Let the diagonals AC and BD of rhombus ABCD intersect at O.

Since the diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90º and AO = CO, BO = OD.

Since ∆AOB is a right triangle right-angle at O.

∴ AB² = OA² + OB²

`AB^2=(1/2AC)^2 +(1/2BD)^2 `

⇒ 4AB² = AC² + BD² ….(i)

Similarly, we have

`4BC^2 = AC^2 + BD^2 ….(ii)`

`4CD^2 = AC^2 + BD^2 ….(iii)`

and,

`4AD^2 = AC^2 + BD^2 ….(iv)`

Adding all these results, we get

`4(AB^2 + BC^2 + AD^2 ) = 4(AC^2 + BD^2 )`

`⇒ AB^2 + BC^2 + AD^2 + DA^2 = AC^2 + BD^2`

Answer 2

In ΔAOB, ΔBOC, ΔCOD, ΔAOD

Applying Pythagoras theorem

AB² = AO² + OB²

BC² = BO² + OC²

CD² = CO² + OD²

AD² = AO² + OD²

Adding all these equations,

AB² + BC² + CD² + AD²

= 2(AO² + OB² + OC² + OD²)

= `2{("AC"/2)^2 + ("BD"/2)^2 + ("AC"/2)^2 + ("BD"/2)^2}`  ...(diagonals bisect each other)

= `2(("AC")^2/2 + ("BD")^2/2)`

= (AC)² + (BD)²