Solution:

In parallelogram ABCD, AB = CD, BC = AD

Draw perpendiculars from C and D on AB as shown.

In right angled ΔAEC, AC^{2} = AE^{2} + CE^{2} [By Pythagoras theorem]

⇒ AC^{2} = (AB + BE)^{2} + CE^{2}

⇒ AC^{2} = AB^{2} + BE^{2} + 2 AB × BE + CE^{2} → (1)

From the figure CD = EF (Since CDFE is a rectangle)

But CD= AB

⇒ AB = CD = EF

Also CE = DF (Distance between two parallel lines)

ΔAFD ≅ ΔBEC (RHS congruence rule)

⇒ AF = BE

Consider right angled ΔDFB

BD^{2} = BF^{2} + DF^{2} [By Pythagoras theorem]

= (EF – BE)^{2} + CE^{2} [Since DF = CE]

= (AB – BE)^{2} + CE^{2} [Since EF = AB]

⇒ BD^{2} = AB^{2} + BE^{2} – 2 AB × BE + CE^{2} → (2)

Add (1) and (2), we get

AC^{2} + BD^{2} = (AB^{2} + BE^{2} + 2 AB × BE + CE^{2}) + (AB^{2} + BE^{2} – 2 AB × BE + CE^{2})

= 2AB^{2} + 2BE^{2} + 2CE^{2}

AC^{2} + BD^{2} = 2AB^{2} + 2(BE^{2} + CE^{2}) → (3)

From right angled ΔBEC, BC^{2} = BE^{2} + CE^{2} [By Pythagoras theorem]

Hence equation (3) becomes,

AC^{2} + BD^{2} = 2AB^{2} + 2BC^{2}

= AB^{2} + AB^{2} + BC^{2} + BC^{2}

= AB^{2} + CD^{2} + BC^{2} + AD^{2}

∴ AC^{2} + BD^{2} = AB^{2} + BC^{2} + CD^{2} + AD^{2}

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