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.
 A: 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 $\theta=15^\circ$.
A: Rotate $D$ around $C$ for $-60^{\circ}$ in to $F$ and let $E$ be intersection point of $AD$ and $BC$. Then $CDEF$ is cyclic and

*

*$ABE$ is congruent to $FBE$ (a.s.a)

*$CDB$ is congruent to $FDB$ (s.a.s) so $\angle CBF = 90^{\circ}$
and so $AEC$ congruent to  $FEC$ (s.a.s). Thus $\angle EAC = \angle EFC = 15^{\circ}$.

A: Reflect $B$ across $AD$ to $F$. Then $\angle FDA = \angle CDA = 30^{\circ}$ so $F,C,D$ are colinear.


*

*Since $\angle CBF = 30^{\circ}$ and $\angle BFC = 75^{\circ}$ we see that $BF = 
    BC=:r$.

*Since $AB = AF$ and $\angle BAF = 60 $ we have $AB = AF = BF =r$.

*So $A,F,C$ are on a circle with center at $B$ and radius $r$, so $$\angle FAC = 
          {1\over 2}\angle CBF = 15^{\circ}$$ and so $ \angle CAD = 15^{\circ}$
A: 
Draw the bisector of $\hat A$, it touches BD at E.Triangle AED is isosceles. Also $\angle DEA=\angle ADC$ that means  CD||AE .$\angle BCE=\angle BAD=30^o$, also $CB \bot AB$, therefore $CE\bot AD$ that is AEDC is a rhombus and triangle ACD is isosceles therefore $\theta=15^o$
A: Here's my solution to this, sorry for being too late.

I'll add some explanation here.

*

*line BK meets AD such that AB=BK=KD.

*Map point B onto point B' and rotate Triangle DCB such that the new triangle DCB' formed it congruent to triangle DCB, and <BDB'=60.

*Join point B' and B, the resulting Triangle BDB' is an equilateral triangle.

*Notice that Triangle BCB' is congruent to Triangle BKD via the ASA property. Therefore, line segment B'C=BC=BK=AB=KD

*Above implies that Triangle CBA is an isosceles right triangle, therefore the unknown angle is 45-30=15.

