Question

In an equilateral triangle ABC, D is a point on the side BC such that

1

BD BC.

3



Prove that 9AD2

= 7AB2

Answer 1

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 .

Answer 2

Answer:

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