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