in an equilateral triangle ABC , …

Given: In an equilateral triangle ΔABC. The side BC is trisected at D such that BD = (1/3) BC. To prove: 9AD2 = 7AB2 Construction: Draw AE ⊥ BC. Proof : In a ΔABC and ΔACE AB = AC ( Given) AE = AE ( common) ∠AEB = ∠AEC = 90° ∴ ΔABC ≅ ΔACE ( For RHS criterion) BE = EC (By C.P.C.T) BE = EC = BC / 2 In a right angled triangle ADE AD2 = AE2 + DE2 ---------(1) In a right angled triangle ABE AB2 = AE2 + BE2 ---------(2) From equ (1) and (2) we obtain ⇒ AD2 - AB2 = DE2 - BE2 . ⇒ AD2 - AB2 = (BE – BD)2 - BE2 . ⇒ AD2 - AB2 = (BC / 2 – BC/3)2 – (BC/2)2 ⇒ AD2 - AB2 = ((3BC – 2BC)/6)2 – (BC/2)2 ⇒ AD2 - AB2 = BC2 / 36 – BC2 / 4 ( In a equilateral triangle ΔABC, AB = BC = CA) ⇒ AD2 = AB2 + AB2 / 36 – AB2 / 4 ⇒ AD2 = (36AB2 + AB2– 9AB2) / 36 ⇒ AD2 = (28AB2) / 36 ⇒ AD2 = (7AB2) / 9 9AD2 = 7AB2 .