In the triangle below
$AD=CD, \: BD=BC$
. The problem is to find a relationship between
$\alpha$
and
$\theta$
.

Label all the angles in terms of
$\alpha$
and
$\theta$
.

Use the sine rule on triangle ABD.
$\frac{sin \alpha}{BD}=\frac{sin(\theta - \alpha )}{AD} \rightarrow \frac{AD}{BD}=\frac{sin (\theta - \alpha )}{sin \alpha}$

and again on triangle BCD.
$\frac{sin (180- \theta )}{CD}=\frac{sin(\theta }{BD} \rightarrow \frac{CD}{BD}=\frac{sin (180 -2 \theta )}{sin \theta}$

$AD=CD$
so
$\frac{sin (\theta - \alpha )}{sin \alpha}=\frac{sin (180 -2\theta )}{sin \theta}$

Use the identities
$sin(\theta - \alpha)=sin \theta cos \alpha - cos \theta sin \alpha$
and
$sin(180-2 \theta )=sin 2 \theta = 2 sin \theta cos \theta$
to get
$\frac{sin \theta cos \alpha - cos \theta sin \alpha}{sin \alpha}= \frac{2 sin \theta cos \theta}{sin \theta}$

$sin \theta cot \alpha - cos \theta = 2 cos \theta$

$sin \theta cot \alpha = 3 cos \theta$

$tan \theta cot \alpha = 3 \rightarrow tan \theta = 3 tan \alpha$