Question: In the diagram ABC is a triangle in which AB = AC and BC = 1. D is the point on BC such that $\angle BAD=\theta$ and $\angle CAD=2\theta$, $BD=x$ and $CD=1-x$.
i) Use the sine rule in each of $\triangle ADB$ and $\triangle ADC$ to show that $cos \theta=\frac{1-x}{2x}$
ii) Hence show that $\frac{1}{3}\lt x \lt \frac{1}{2}$
This is the diagram that I'm working off.
This is for the Proofs topic of the Extension 2 HSC maths course. This belongs to a section introducing the Ineqaulity proofs.
I have successfully done part i)
$\frac{sin\theta}{x}=\frac{sin\mu}{AD}$, $\frac{sin2\theta}{1-x}=\frac{sin\mu}{AD}$
$\frac{sin\theta}{x}=\frac{sin2\theta}{1-x}$
$\frac{sin\theta}{x}=\frac{2sin\theta cos\theta}{1-x}$
$\frac{1-x}{2xcos\theta}=1$
$\therefore cos\theta=\frac{1-x}{2x}$
But part ii) really baffles me. Any help would be greatly appreciated. Thank you!
For ii), we have $ 0< \theta < \frac \pi3\implies \frac12<\cos\theta = \frac{1-x}{2x} <1$, which leads to
$$\frac{1}{3}\lt x \lt \frac{1}{2}$$