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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. - Geometry

ConceptApollonius Theorem

#### Question

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.

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#### APPEARS IN

Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 2: Pythagoras Theorem
Problem Set 2 | Q: 9 | Page no. 45
Solution Prove that the Sum of the Squares of the Diagonals of a Parallelogram is Equal to the Sum of the Squares of Its Sides. Concept: Apollonius Theorem.
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