Q28 of 32 Page 6

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. (CBSE 2005, 06, 08)

In Rhombus ABCD, 

AB, BC, CD and AD are the sides of the rhombus.
BD and AC are the diagonals.

To prove: AB2 + BC2 + CD2 +AD2 = AC2 + BD2

Proof:
The figure is shown below:

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain


AB2 = AO2 + OB2 .....................eq(i)


BC2 = BO2 + OC2  .......................eq (ii)


CD2 = CO2 + OD2 ....................eq(iii)


AD2 = AO2 + OD2.....................eq (iv)


Now after adding all equations, we get,


AB2 + BC2 + CD2 + AD2 = 2 (AO2 + OB2 + OC2 + OD2)

Diagonals of a rhombus bisect each other,
Thus AO = AC/2, OB = BD/2, OC=AC/2, and OD= BD/2


=(AC)2 + (BD)2

Hence Sum of squares of sides of a rhombus equals to sum of squares of diagonals of rhombus.

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