# 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

2015-05-29T22:05:34+05:30

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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...

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