Question

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

Answer 1

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.