Figure

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 = BE

Consider right angled ΔDFB

BD²= 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+ BC2² = 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.