Question

In equilateral ΔABC, D ∈ BC such that BD: DC = 1:2. Prove that 3AD=√7 AB.

Answer 1

Steps:
1) Let 'x'be such that '3x'is side of triangle and
$BD = x \\ CD = 2x$

2) We take A as origin and Mark co-ordinates of other vertices by properties of triangle.
BC is base,
=>D = (x, 0)
& => C = (3x, 0)

3) By Point -Form,
$A = (3x \cos(60\degree) , 3x \sin(60\degree ) ) \\ => A = ( \frac{3x}{2}, \frac{3 \sqrt{3}}{2} )$

4) By Distance Formula,
$AD = \sqrt{(3x/2 -x )^2, (3\sqrt{3}x/2 -0 )^2} \\ => AD = \sqrt{7} \: units$

We know,
AB = 3x ( Side of an equilateral triangle)

5) Then, we can say
$\boxed{\frac{AD}{AB} = \frac{\sqrt{7}}{3} }$

Hence Proved :