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prove that the sum of the square of the diagonals of parallelogram is equal to the sum of square of its sides​

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Asked by Vaishali mathur Mathur

Aug 18, 2013

Prove that the sum of the squares of the diagonals of parallelogram

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

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person

Kishore Kumar

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 + BD2 = (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 + BC2 + BC2 = AB2 + CD2 + BC2 + AD2 ∴ AC2 + BD2 = AB2 + BC2 + 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.