<s>[INST] Find $\angle CAD$ if $\triangle ABC$ is right angled at $B$, $\angle BAD = 30^\circ, \angle ADB = \angle ADC = 15^\circ$ Find angle $\theta$ in the below diagram.

This is a question that was brought to me by a high school student.
While I came up with a trigonometric solution and a synthetic solution, I am posting here to see more solutions that others come up with (esp. other synthetic solutions)
My immediate solution involved combination of a simple construction and trigonometry, basically knowing that $\tan 30^\circ = \dfrac{1}{\sqrt3}$ and $\tan 15^\circ = 2 - \sqrt3$.

We draw perp from $D$ to $AB$ extend and $BC$ extend. We also note that $\angle DBC = \angle DBE = 45^\circ$. If $AB = x, BE = y$, we find $x$ in terms of $y$. We next find $CF$ in terms of $y$ and subtracting from $y$ gives us $BC$ and we show $BC = x$.
Then in search of a synthetic solution, I drew a few more lines and as $FE$ is perpendicular bisector of $BD$, $BG = GD, BH = HD$.

So we see that $AB = BG$ and by A-A-S, $\triangle BHC \cong \triangle BHG$ which leads to $AB = BC$ and we have $\theta = 15^\circ$.
Look forward to more interesting solutions.
 [/INST] Sine rule for $\Delta ABD$:
$$\frac{AB}{\sin 15^\circ}=\frac{BD}{\sin 30^\circ} \Rightarrow BD=\frac{AB}{2\sin 15^\circ}$$
Sine rule for $\Delta BCD$:
$$\frac{BC}{\sin 30^\circ}=\frac{BD}{\sin 105^\circ} \Rightarrow BC=\frac{BD}{2\sin 105^\circ}=\frac{BD}{2\cos 15^\circ}=\frac{AB}{4\sin 15^\circ\cos 15^\circ}=AB$$
Hence, the triangle $ABC$ is right angled and isosceles, implying $\theta=15^\circ$.
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