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

Std 10 Gujarat Board

1
by dkdk

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Std 10 Gujarat Board

by dkdk

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See diagram...

we apply the principle of Pythagoras..

BE² + DE² = BD² --- (1)

AF² + CF² = AC² --- (2)

BE² = AB² - AE²

CF² = CD² - DF² so substituting them in the equations above,

AC² + BD²

= AB² + CD² + DE² + AF² - AE² - DF²

= AB² + CD² + (AD - AE)² + (AD + DF)² - AE² - DF²

= AB² + CD² + 2 AD² + AE² - 2 AD AE + DF² + 2 AD DF - AE² - DF²

= AB² + CD² + AD² + CB² + 2 AD (DF - AE) as CB = AD..

In triangles ABE and CDF, the angles BAE and CDF are equal.. AB = CD. BE and CF are equal , as they are distance between parallel lines. Hence, AE = DF as they are two congruent triangles...

Thus, DF = AE..

So, sum of squares of diagonals = sum of squares of sides...

we apply the principle of Pythagoras..

BE² + DE² = BD² --- (1)

AF² + CF² = AC² --- (2)

BE² = AB² - AE²

CF² = CD² - DF² so substituting them in the equations above,

AC² + BD²

= AB² + CD² + DE² + AF² - AE² - DF²

= AB² + CD² + (AD - AE)² + (AD + DF)² - AE² - DF²

= AB² + CD² + 2 AD² + AE² - 2 AD AE + DF² + 2 AD DF - AE² - DF²

= AB² + CD² + AD² + CB² + 2 AD (DF - AE) as CB = AD..

In triangles ABE and CDF, the angles BAE and CDF are equal.. AB = CD. BE and CF are equal , as they are distance between parallel lines. Hence, AE = DF as they are two congruent triangles...

Thus, DF = AE..

So, sum of squares of diagonals = sum of squares of sides...