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# Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. Q.6

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The best question of optional chapter of class 10th math of chapter name Triangles of exercise 6.6, Give me the best way to find this question Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides

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1. Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.

By applying Pythagoras Theorem in âˆ†DEA, we get,

DE2 + EA2 = DA2 â€¦â€¦â€¦â€¦â€¦â€¦.â€¦ (i)

By applying Pythagoras Theorem in âˆ†DEB, we get,

DE2 + EB2 = DB2

DE2 + (EA + AB) 2 = DB2

(DE2 + EA2) + AB2 + 2EA Ã— AB = DB2

DA2 + AB2 + 2EA Ã— AB = DB2 â€¦â€¦â€¦â€¦â€¦. (ii)

By applying Pythagoras Theorem in âˆ†ADF, we get,

AD2 = AF2 + FD2

Again, applying Pythagoras theorem in âˆ†AFC, we get,

AC2 = AF2 + FC2 = AF2 + (DC âˆ’ FD) 2

= AF2 + DC2 + FD2 âˆ’ 2DC Ã— FD

= (AF2 + FD2) + DC2 âˆ’ 2DC Ã— FD AC2

AC2= AD2 + DC2 âˆ’ 2DC Ã— FD â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (iii)

Since ABCD is a parallelogram,

AB = CD â€¦â€¦â€¦â€¦â€¦â€¦â€¦.â€¦(iv)

And BC = AD â€¦â€¦â€¦â€¦â€¦â€¦. (v)

In âˆ†DEA and âˆ†ADF,

âˆ DEA = âˆ AFD (Each 90Â°)

âˆ EAD = âˆ ADF (EA || DF)

âˆ´ âˆ†EAD â‰… âˆ†FDA (AAS congruence criterion)

â‡’ EA = DF â€¦â€¦â€¦â€¦â€¦â€¦ (vi)

Adding equations (i) and (iii), we get,

DA2 + AB2 + 2EA Ã— AB + AD2 + DC2 âˆ’ 2DC Ã— FD = DB2 + AC2

DA2 + AB2 + AD2 + DC2 + 2EA Ã— AB âˆ’ 2DC Ã— FD = DB2 + AC2

From equation (iv) and (vi),

BC2 + AB2 + AD2 + DC2 + 2EA Ã— AB âˆ’ 2AB Ã— EA = DB2 + AC2

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

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