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