ABCD is a rhombus. Prove that AB^{2} + BC^{2} + CD^{2} + DA^{2}= AC^{2} + BD^{2}

#### Solution 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.

∴ AB^{2} = OA^{2} + OB^{2}

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

⇒ 4AB^{2} = AC^{2} + BD^{2} ….(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`

#### Solution 2

In ΔAOB, ΔBOC, ΔCOD, ΔAOD

Applying Pythagoras theroem

AB^{2} = AD^{2} + OB^{2}

BC^{2} = BO^{2} + OC^{2}

CD^{2} = CO^{2} + OD^{2}

AD^{2} = AO^{2} + OD^{2}

Adding all these equations,

AB^{2} + BC^{2} + CD^{2} + AD^{2}

= 2(AD^{2} + OB^{2} + OC^{2 }+ OD^{2})

= `2(("AC"/2)^2 + ("BD"/2)^2 + ("AC"/2)^2 + ("BD"/2)^2)` ...(diagonals bisect each othar.)

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

= (AC)^{2} + (BD)^{2}.