Question

In an equilateral triangle ABC, the side BC is trisected at D, then 9AD² is .......

(1) 7AB² 

(2) 8BC² 

(3) 4AC²

(4) 3/2 AB²

Answer 1

Correct option is (1) 7AB²

Let A(0, 0), B(2a, 0), C(a, √3 a),

Then D \((\frac{4a}{3},\frac{2\sqrt{3}}{3})\)

⇒ 9 AD² = 28a² 

= 7 AB²