Prove that sum of squares of diagonals of parallelogram are equal to sum of squares of its sides

Prove that sum of squares of diagonals of parallelogram are equal to s

1.	Prove that sum of squares of diagonals of parallelogram are equal to sum of squares of its sides
Answer» Here ,a short ans for this question we have to prove that ab2+bc2+cd2+da2=ac2+bd2 since In parallelogram ABCD, AB = CD, BC = ADDraw perpendiculars from C and D on AB as shown.In right angled ΔAEC, AC2\xa0= AE2\xa0+ CE2\xa0[By Pythagoras theorem]⇒ AC2\xa0= (AB + BE)2\xa0+ CE2⇒ AC2\xa0= AB2\xa0+ BE2\xa0+ 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\xa0= BF2\xa0+ DF2\xa0[By Pythagoras theorem] = (EF – BE)2\xa0+ CE2 [Since DF = CE] = (AB – BE)2\xa0+ CE2 [Since EF = AB]\xa0⇒ BD2\xa0= AB2\xa0+ BE2\xa0– 2 AB × BE + CE2 → (2)Add (1) and (2), we getAC2\xa0+ BD2\xa0= (AB2\xa0+ BE2\xa0+ 2 AB × BE + CE2) + (AB2\xa0+ BE2\xa0– 2 AB × BE + CE2) = 2AB2\xa0+ 2BE2\xa0+ 2CE2 AC2\xa0+ BD2\xa0= 2AB2\xa0+ 2(BE2\xa0+ CE2) → (3)From right angled ΔBEC, BC2\xa0= BE2\xa0+ CE2\xa0[By Pythagoras theorem]Hence equation (3) becomes, AC2\xa0+ BD2\xa0= 2AB2\xa0+ 2BC2 = AB2\xa0+ AB2\xa0+ BC2\xa0+ BC2 = AB2\xa0+ CD2\xa0+ BC2\xa0+ AD2∴ AC2\xa0+ BD2\xa0= AB2\xa0+ BC2\xa0+ CD2\xa0+ AD2Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.Thank you?