Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
ABCD is a rhombus whose diagonals AC and BD intersect at O. [Given]
We have to prove that,
AB2 + BC2 + CD2 + AD2 = AC2 + BD2
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB2 = AO2 + BO2 ... (i) [By Pythagoras]
Similarly,
AD2 = AO2 + DO2 ... (ii)
DC2 = DO2 + CO2 ... (iii)
BC2 = CO2 + BO2 ... (iv)
Adding equations (i) + (ii) + (iii) + (iv) we get,
AB2 + AD2 + DC2 + BC2 = 2(AO2 + BO2 + DO2 + CO2 )
= 4AO2 + 4BO2 [Since, AO = CO and BO =DO]
= (2AO)2 + (2BO)2 = AC2 + BD2
