Question

prove that sum of squares of sides of a parallelogram is equal to sum of squares of its diagonals.

Answer 1

Let ABCD be aparallelogram.

Let us drawperpendicular DE on extended side AB, and AF on side DC.

Applying Pythagorastheorem in ΔDEA,we obtain

DE2+ EA2= DA2 …(i)

Applying Pythagorastheorem in ΔDEB,we obtain

DE2+ EB2= DB2

DE2+ (EA + AB)2= DB2

(DE2+ EA2)+ AB2+ 2EA × AB = DB2

DA2+ AB2+ 2EA × AB = DB2 …(ii)

Applying Pythagorastheorem in ΔADF, we obtain

AD2 = AF2+ FD2

Applying Pythagorastheorem in ΔAFC, we obtain

AC2 = AF2+ FC2

=AF2 + (DC − FD)2

=AF2 + DC2 + FD2 − 2DC ×FD

=(AF2 + FD2) + DC2 − 2DC ×FD

AC2 = AD2+ DC2 − 2DC × FD … (iii)

Since ABCD is aparallelogram,

AB = CD …(iv)

And, BC = AD …(v)

In ΔDEAand ΔADF,

∠DEA =∠AFD (Both 90°)

∠EAD = ∠ADF (EA|| DF)

AD = AD (Common)

∴ ΔEAD is congruent to ΔFDA (AAS congruencecriterion)

⇒ EA = DF …(vi)

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

DA2+ AB2+ 2EA × AB + AD2+ DC2− 2DC × FD = DB2+ AC2

DA2+ AB2+ AD2+ DC2+ 2EA × AB − 2DC × FD = DB2+ AC2

BC2 + AB2+ AD2 + DC2 + 2EA × AB − 2AB ×EA = DB2 + AC2

[Using equations (iv)and (vi)]

AB2+ BC2+ CD2+ DA2= AC2+ BD2



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Answer 2

Answer:

Step-by-step explanation:

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