Data: ABCD is a rhombus.
Here, AB = BC = CD = DA.
Diagonals AC and BD intersects at ’O’.
To Prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2 .
In rhombus diagonals bisects perpendicularly.
∴ ∠AOB = ∠AOD = 90°.
In ⊥∆AOB,
AB2 = AO2 + BO2 ……..(i)
In ⊥∆BOC,
AC2 = BO2 + CO2 ……..(ii)
In ⊥∆COD,
CD2 = OC2 + OD2 ……….(iii)
In ⊥∆AOD,
AD2 = AO2 + OD2…………(iv)
By Adding equations (i) + (ii) + (iii) + (iv)
AB2 + BC2 + CD2 + DA2
= AO2 + BO2 + BO2 + CO2 + CO2 + DO2 + AO2 + OD2
= 2AO2 + 2BO2 + 2CO2 + 2DO2
= 2AO2 + 2CO2 + 2BO2 + 2DO2
Now, RHS = 2AO2 + 2CO2 + 2BO2 + 2DO2
RHS = AC2 + BD2
∴ LHS = RHS
∴ AB2 + BC2 + CD2 + DA2
= AC2 + BD2 .
