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

**Solution:**

Let's draw a parallelogram ABCD as shown below,

AB = CD, AD = BC [Since opposite sides of a parallelogram are equal]

AC and BD are the diagonals of the parallelogram ABCD

Draw AE ⊥ CD, DF ⊥ AB

We have to prove that, AC² + BD² = AB² + BC² + CD² + AD²

EA = DF (Perpendiculars drawn between same parallel lines)

According to Pythagoras theorem,

In ΔAEC

AC² = AE² + EC²

AC² = AE² + (ED + DC)² [Since EC = ED + DC]

AC² = AE² + DE² + DC² + 2DE.DC [Since (a + b)² = a² + b² + 2ab]

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

In ΔDFB

BD² = DF² + BF²

BD² = DF² + (AB - AF)² [Since BF = AB = AF]

BD² = DF² + AB² + AF² - 2AB.AF [Since (a - b)² = a² + b² - 2ab]

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

Adding (i) and (ii)

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

AC² + BD² = BC² + DC² + 2AB.AF + AD² + AB² - 2AB.AF [Since AD = BC and DE = AF, CD = AB]

Thus, AC² + BD² = AB² + BC² + CD² + AD²

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**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

**Summary:**

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.

**☛ Related Questions:**

- In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that: (i) ΔAPC ~ ΔDPB (ii) AP.PB = CP.DP.
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- In Fig. 6.63, D is a point on side BC of ∆ABC such that BD/CD = BA/CA. Prove that: AD is the bisector of ∠BAC.
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