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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# 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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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,

DE

^{2}+ EA^{2}= DA^{2}â€¦â€¦â€¦â€¦â€¦â€¦.â€¦ (i)By applying Pythagoras Theorem in âˆ†DEB, we get,

DE

^{2}+ EB^{2}= DB^{2}DE

^{2}+ (EA + AB)^{ 2}= DB^{2}(DE

^{2}+ EA^{2}) + AB^{2}+ 2EA Ã— AB = DB^{2}DA

^{2}+ AB^{2}+ 2EA Ã— AB = DB^{2}â€¦â€¦â€¦â€¦â€¦. (ii)By applying Pythagoras Theorem in âˆ†ADF, we get,

AD

^{2}= AF^{2}+ FD^{2}Again, applying Pythagoras theorem in âˆ†AFC, we get,

AC

^{2}= AF^{2}+ FC^{2}= AF^{2}+ (DC âˆ’ FD)^{ 2}= AF

^{2}+ DC^{2}+ FD^{2}âˆ’ 2DC Ã— FD= (AF

^{2}+ FD^{2}) + DC^{2}âˆ’ 2DC Ã— FD AC^{2}AC

^{2}= AD^{2}+ DC^{2}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)

AD = AD (Common Angles)

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,

DA

^{2}+ AB^{2}+ 2EA Ã— AB + AD^{2}+ DC^{2}âˆ’ 2DC Ã— FD = DB^{2}+ AC^{2}DA

^{2}+ AB^{2}+ AD^{2}+ DC^{2}+ 2EA Ã— AB âˆ’ 2DC Ã— FD = DB^{2}+ AC^{2}From equation (iv) and (vi),

BC

^{2}+ AB^{2}+ AD^{2}+ DC^{2}+ 2EA Ã— AB âˆ’ 2AB Ã— EA = DB^{2}+ AC^{2}AB

^{2}+ BC^{2}+ CD^{2}+ DA^{2}= AC^{2}+ BD^{2}