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

Chapter 2 – Pythagoras Theorem- Text Book Solution

Problem Set 2 | Q 9 | Page 45
Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

solution

Diagonals of a parallelogram bisect each other.
i.e. O is the mid point of AC and BD.
In ∆ABD, point O is the midpoint of side BD.

B O = O D = \frac{1}{2} B D

{A B}^{2} + {A D}^{2} = 2 {A O}^{2} + 2 {B O}^{2} (by Apollonius theorem) . . . (1)

In ∆CBD, point O is the midpoint of side BD.

B O = O D = \frac{1}{2} B D

{C B}^{2} + {C D}^{2} = 2 {C O}^{2} + 2 {B O}^{2} (by Apollonius theorem) . . . (2)

Adding (1) and (2), we get

${A B}^{2} + {A D}^{2} + {C B}^{2} + {C D}^{2} = 2 {A O}^{2} + 2 {B O}^{2} + 2 {C O}^{2} + 2 {B O}^{2}$
$\Rightarrow {A B}^{2} + {A D}^{2} + {C B}^{2} + {C D}^{2} = 2 {A O}^{2} + 4 {B O}^{2} + 2 {A O}^{2} (∵ O C = O A)$
$\Rightarrow {A B}^{2} + {A D}^{2} + {C B}^{2} + {C D}^{2} = 4 {A O}^{2} + 4 {B O}^{2}$
$\Rightarrow {A B}^{2} + {A D}^{2} + {C B}^{2} + {C D}^{2} = {(2 A O)}^{2} + {(2 B O)}^{2}$
$\Rightarrow {A B}^{2} + {A D}^{2} + {C B}^{2} + {C D}^{2} = {(A C)}^{2} + {(B D)}^{2}$
$\Rightarrow {A B}^{2} + {A D}^{2} + {C B}^{2} + {C D}^{2} = {A C}^{2} + {B D}^{2}$

Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Explanation:-

Let ABCD be a parallelogram with diagonals AC and BD intersecting at point O.

Using the Law of Cosines, we have:

AC² = AB² + BC² – 2AB x BC cos(∠ABC) BD² = BC² + CD² – 2BC x CD cos(∠BCD)

Since ABCD is a parallelogram, we have AB = CD and BC = AD. Substituting these values in the above equations, we get:

AC² = AB² + BC² – 2AB x BC cos(∠ABC) BD² = BC² + CD² – 2BC x CD cos(∠BCD) AC² + BD² = 2AB² + 2BC²

Also, we have:

OA² = OB² = AB² + BO² OC² = OD² = CD² + CO²

Adding these equations, we get:

OA² + OC² + 2AB² + 2BC² + BO² + CO² = AC² + BD²

Since AB = CD and BC = AD, we have BO = CO and 2AB² + 2BC² = 4AB². Substituting these values, we get:

OA² + OC² + 4AB² = AC² + BD²

This proves that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Chapter 2 – Pythagoras Theorem- Text Book Solution

Problem Set 2 | Q 9 | Page 45