Geometry

$\color{blac}Find\ \angle CAD\ in\ degrees.\\$

Thanks webbinsight for the useful suggestions to solve this tricky task!

$\angle ADB =\angle ADE=\alpha\\ \angle ACB=\angle ACE=\beta$

The angles intersection AD/EC are twice:

$(120°-\alpha )\ and\ (60°+\alpha )$

The angles intersection AC/BD are twice:

$(30°+\beta)\ and\ (150°-\beta)\\ \angle CAD=60°+\alpha -\beta$

In the triangle (intersection AC/BD) - A-D the following applies:

$\alpha =180°-(150°-\beta )-(60°+\alpha -\beta)\\ \color{blue}\beta = 15°$

The angles intersection BD/CE are twice:

$(180°-30°-2\beta=120°)\ and\ (60°)$

In the triangle (intersection AD/CE) -  (intersection CE/BD) - D the following applies:

$(120°)+(60°+\alpha )+(\alpha )=180°\\ \color{blue} \alpha =0\ ?\ ?\ ?$

I ask everyone for help.

 !