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

Let ABCD be a parallelogram, with AB = CD ; AB || CD and BC = AD ; BC || AD.

Construct AE ⊥ CD and extend CD to F such that, BF ⊥ CF.

In ΔAED and ΔBCF

AE = BF [Distance between two parallel lines i.e. AB and CD]

AD = BC [opposite sides of a parallelogram are equal]

∠AED = ∠BFC [Both 90°]

ΔAED ≅ ΔBCF [By Right Angle - Hypotenuse - Side Criteria]

⇒ DE = CF [Corresponding sides of congruent triangles are equal] [1]

In ΔBFD, By Pythagoras theorem i.e.

(Hypotenuse)² = (base)² + (Perpendicular)²

BD² = DF² + BF²

⇒ BD² = (CD + CF)² + BF² [2]

In ΔAEC, By Pythagoras theorem

AC² = AE² + CE²

⇒ AC² = AE² + (CD - AE)²

⇒ AC² = BF² + (CD - CF)² [As, AE = BF and CF = AE] [2]

In ΔBCF, By Pythagoras theorem,

BC² = BF² + CF²

BF² = BC² - CF² [3]

Adding [2] and [3]

BD² + AC² = 2BF² + (CD + CF)² + (CD - CF)²

⇒ BD² + AC²= 2BC² - 2CF² + CD² + CF² + 2CD.CF + CD² + CF² - 2CD.CF

⇒ BD² + AC² = 2BC² + 2CD²

⇒ BD² + AC² = BC² + BC² + CD² + CD²

⇒ BD² + AC² = AB² + BC² + CD² + AD² [since BC = AD and AB = CD]

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