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

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In parallelogram ABCD, AB = CD, BC = AD
Draw perpendiculars from C and D on AB as shown.
In right angled ΔAEC, AC2 = AE2 + CE2 [By Pythagoras theorem]
⇒ AC2 = (AB + BE)2 + CE2
⇒ AC2 = AB2 + BE2 + 2 AB × BE + CE2  → (1)
From the figure CD = EF (Since CDFE is a rectangle)
But CD= AB
⇒ AB = CD = EF
Also CE = DF (Distance between two parallel lines)
ΔAFD ≅ ΔBEC (RHS congruence rule)
⇒ AF = BE
Consider right angled ΔDFB
BD2 = BF2 + DF2 [By Pythagoras theorem]
        = (EF – BE)2 + CE2  [Since DF = CE]
        = (AB – BE)2 + CE2   [Since EF = AB]
 ⇒ BD2 = AB2 + BE2 – 2 AB × BE + CE2  → (2)
Add (1) and (2), we get
AC2 + BD= (AB2 + BE2 + 2 AB × BE + CE2) + (AB2 + BE2 – 2 AB × BE + CE2)
                     = 2AB2 + 2BE2 + 2CE2
  AC2 + BD2 = 2AB2 + 2(BE2 + CE2)  → (3)
From right angled ΔBEC, BC2 = BE2 + CE2 [By Pythagoras theorem]
Hence equation (3) becomes,
  AC2 + BD2 = 2AB2 + 2BC2
                                  = AB2 + AB2 + BC+ BC2
                                  = AB2 + CD2 + BC+ AD2
∴   AC2 + BD= AB2 + BC+ CD2 + AD2
Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
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