Relation Between Angles in Adjacent Triangles Problem

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

trigonometry problem

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

trigonometry problem

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
\[\frac{sin \theta cos \alpha - cos \theta sin \alpha}{sin \alpha}= \frac{2 sin \theta cos \theta}{sin \theta}\]

\[sin \theta cot \alpha - cos \theta = 2 cos \theta\]

\[sin \theta cot \alpha = 3 cos \theta\]

\[tan \theta cot \alpha = 3 \rightarrow tan \theta = 3 tan \alpha\]