Prove that the sum of the …

Answer, 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.

Dont ask maths

Use appoloneous theorem