Prove $\tan \theta = 3 \tan \alpha$ Prove $\tan \theta = 3 \tan \alpha$


The question is related to this diagram.

I tried approaching this problem using the cosine and sine rules for the triangles in this figure. I was hoping that if I could find $\sin \theta, \sin \alpha, \cos \theta$ and $\cos \alpha$, I could try to simplify $ \tan \theta$ to get $3\tan \alpha$. However, that didn't work. I ended up with

$\tan \theta = \frac{-BD^2}{BC^2}\cdot  \sin 2\theta $, which doesn't seem to be helpful. Any hints or answers would be appreciated.
It is obvious that $\bigtriangleup BDC$ is isoscales triangle. So, if you mark the midpoint of the side $\overline{BC}$ with $E$, you will have right triangle $\bigtriangleup BDE$, from which you can find: $$\tan \theta = \frac{|\overline{DE}|}{|\overline{BE}|}.$$

Next, your triangle $\bigtriangleup ADE$ is also right triangle, so it holds: 
$$\tan \alpha = \frac{|\overline{DE}|}{|\overline{AE}|}.$$

Since $\quad \overline{AE} = 3\overline{BE} \quad$ we have:
$$\tan \alpha = \frac{|\overline{DE}|}{3|\overline{BE}|}.$$

Finally, multiplying with $3$ will result with:
$$3 \tan \alpha = \tan \theta.$$
By the law of sines for $\Delta ADC$ we obtain: $$\frac{AC}{\sin(\alpha+\theta)}=\frac{DC}{\sin\alpha}.$$
Also, $$AC=2BC=4DC\cos\theta.$$
Thus, $$4\sin\alpha\cos\theta=\sin(\alpha+\theta)$$ or
$$4\sin\alpha\cos\theta=\sin\alpha\cos\theta+\cos\alpha\sin\theta$$ or

$$3\sin\alpha\cos\theta=\cos\alpha\sin\theta$$ or
$$3\tan\alpha=\tan\theta.$$