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

**Solution:**

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Theorem 6.8.

In parallelogram ABCD

AB = CD AD = BC

Draw AE ⊥ CD, DF ⊥ AB

EA = DF (Perpendiculars drawn between same parallel lines)

In ΔAEC

AC² = AE² + EC²

= AE² +[ED + DC]²

= AE² + DE² + DC² + 2DE.DC

AC² = AD² + DC² + 2DE × DC .....(i) [Since, AD² = AE² + DE²]

In ΔDFB

BD² = DF² + BF²

= DF² +[ AB - AF]²

= DF² + AB² + AF² - 2 AB.AF

= AD² + AB² - 2 AB.AF

BD² = AD² + AB² - 2 AB.AF .....(ii) [Since, AD² = DF² + AF²]

Adding (i) and (ii)

AC² + BD² = AD² + DC² + 2DE.DC + AD² + AB² - 2 AB.AF

AC² + BD² = BC² + DC² + 2AB.AF + AD² + AB² - 2AB.AF

(Since AD = BC and DE = AF, CD = AB)

⇒ AC² + BD² = AB² + BC² + CD² + AD²

**Video Solution:**

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

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.6 Question 6:

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

Hence proved that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides