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

81.6k views

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

by (9.9k points)
selected

Solution: In parallelogram ABCD, AB = CD, BC = AD
Draw 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 = EF
Also 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 get
AC2 + 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 + AD2
Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.