Question

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

Answer 1

In parallelogram ABCD, AB = CD, BC = ADDraw 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

Answer 2

Answer:

Step-by-step explanation:

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