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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.

\[BO = OD = \frac{1}{2}BD\]

\[{AB}^2 + {AD}^2 = 2 {AO}^2 + 2 {BO}^2 \left( \text{by Apollonius theorem} \right) . . . \left( 1 \right)\]

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

\[BO = OD = \frac{1}{2}BD\]

\[{CB}^2 + {CD}^2 = 2 {CO}^2 + 2 {BO}^2 \left( \text{by Apollonius theorem} \right) . . . \left( 2 \right)\]

Adding (1) and (2), we get

\[{AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = 2 {AO}^2 + 2 {BO}^2 + 2 {CO}^2 + 2 {BO}^2 \]
\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = 2 {AO}^2 + 4 {BO}^2 + 2 {AO}^2 \left( \because OC = OA \right)\]
\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = 4 {AO}^2 + 4 {BO}^2 \]
\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = \left( 2AO \right)^2 + \left( 2BO \right)^2 \]
\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = \left( AC \right)^2 + \left( BD \right)^2 \]
\[ \Rightarrow {AB}^2 + {AD}^2 + {CB}^2 + {CD}^2 = {AC}^2 + {BD}^2\]

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