TITLE: Find $\angle CAD$ if $\triangle ABC$ is right angled at $B$, $\angle BAD = 30^\circ, \angle ADB = \angle ADC = 15^\circ$
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: 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}$