Show that $|BD| = 2y \tan 2\theta$

Question

In the triangle $ABC$, the angle at $C$ is $90^o$, $|AC|=y$.

$D$ is a point on $[BC]$.

Angle $BAD=2\theta$, and angle $DAC=45^o -\theta$.

$0^o \lt \theta \lt 45^o$

Show that $|BD|=2y\tan 2\theta$

I've been looking at this for a few days and can't figure it out.

It is a question in a section on double-angle formulae so I started with:

$\tan(2\theta+(45^o -\theta))=\frac{|BD|+|DC|}{y}=\frac{\tan 2\theta +\frac{|DC|}{y}}{1-(\tan 2\theta)(\frac{|DC|}{y})}$

Then I tried starting with:

$\tan (\theta +45^o)=\frac{\tan \theta +1}{1-\tan \theta}$

I tried tidying up the expressions but to no avail. Any ideas?

Answer 1

HINT

By rule of sine we have

$$\frac{|BD|}{\sin 2\theta}=\frac{|DA|}{\sin (45°-\theta)}$$

and $|DA|$ can be related to $|AC|=y$.