Prove that sum of squares of …

Here ,a short ans for this question we have to prove that ab2+bc2+cd2+da2=ac2+bd2 since

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. Thank you?