In parallelogram ABCD, AB = CD, BC = ADDraw perpendiculars from C and D on AB as shown.In right angled ΔAEC, AC2= AE2+ CE2[By Pythagoras theorem]⇒ AC2= (AB + BE)2+ CE2⇒ AC2= AB2+ BE2+ 2 AB × BE + CE2 → (1)From the figure CD = EF (Since CDFE is a rectangle)But CD= AB⇒ AB = CD = EFAlso CE = DF (Distance between two parallel lines)ΔAFD ≅ ΔBEC (RHS congruence rule)⇒ AF = BEConsider right angled ΔDFBBD2= BF2+ DF2[By Pythagoras theorem] = (EF – BE)2+ CE2[Since DF = CE] = (AB – BE)2+ CE2 [Since EF = AB]⇒ BD2= AB2+ BE2– 2 AB × BE + CE2 → (2)Add (1) and (2), we getAC2+ BD2= (AB2+ BE2+ 2 AB × BE + CE2) + (AB2+ BE2– 2 AB × BE + CE2) = 2AB2+ 2BE2+ 2CE2 AC2+ BD2= 2AB2+ 2(BE2+ CE2) → (3)From right angled ΔBEC, BC2= BE2+ CE2[By Pythagoras theorem]Hence equation (3) becomes, AC2+ BD2= 2AB2+ 2BC2 = AB2+ AB2+ BC2+ BC2 = AB2+ CD2+ BC2+ AD2∴ AC2+ BD2= AB2+ BC2+ CD2+ AD2Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides