#### Question

Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

#### Solution

Diagonals of a parallelogram bisect each other.

i.e. O is the mid point of AC and BD.

In ∆ABD, point O is the midpoint of side BD.

In ∆CBD, point O is the midpoint of side BD.

Adding (1) and (2), we get

\[{AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = 2 {AO}^2 + 2 {BO}^2 + 2 {CO}^2 + 2 {BO}^2 \]

\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = 2 {AO}^2 + 4 {BO}^2 + 2 {AO}^2 \left( \because OC = OA \right)\]

\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = 4 {AO}^2 + 4 {BO}^2 \]

\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = \left( 2AO \right)^2 + \left( 2BO \right)^2 \]

\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = \left( AC \right)^2 + \left( BD \right)^2 \]

\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = {AC}^2 + {BD}^2\]

Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.