Let A B C be a triangle and let D be the mid-point of B C . Suppose cot ∠ C A D : cot ∠ B A D = 2 : 1 . If G is the centroid of Δ A B C , then the measure of ∠ B G A is

Step-by-step solution

Let Δ A B C and D is the mid-point of side B C , then According to Apollonius theorem, A D 2 = 1 2 b 2 + c 2 - a 2 4 ⇒ A D = 1 2 2 b 2 + 2 c 2 - a 2 ∵ A G G D = 2 1 ⇒ A G = 2 3 A D = 1 3 2 b 2 + 2 c 2 - a 2 Similarly B G = 1 3 2 c 2 + 2 a 2 - b 2 So, cos θ = B G 2 + A G 2 - A B 2 2 B G A G , where θ = ∠ A G B = 1 9 2 c 2 + 2 a 2 - b 2 + 1 9 2 b 2 + 2 c 2 - a 2 - c 2 2 B G A G = a 2 + b 2 - 5 c 2 18 B G A G . . . ( i ) Now, as cot θ 2 = 2 cot θ 1 ⇒ cos θ 2 sin θ 2 = 2 cos θ 1 sin θ 1 ⇒ A D 2 + b 2 - a 2 4 2 A D · b