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.
Best Answer
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: $$3 \tan \alpha = \tan \theta.$$