Question

Pls answer the circled one...

Answer 1

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 .

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