You'll be stuck

Let $BC = a$ , $AC = b$ , $AB = c$ , $AD = d$ and $\angle BAD = \theta$ . We note that $\angle BAC = \theta + 15^\circ$ . Using Sine Rule, we have:

$\begin {cases} \dfrac {\sin{\theta}}{\frac{a}{2}} = \dfrac {\sin {45^\circ}}{c} & \Rightarrow \sin {\theta} = \dfrac {a}{2\sqrt{2}c} \\ \dfrac {\sin{(\theta+15^\circ)}}{a} = \dfrac {\sin {30^\circ}}{c} & \Rightarrow \sin {(\theta+15^\circ)} = \dfrac {a}{2c} \end {cases}$

$\Rightarrow \dfrac { \sin {(\theta+15^\circ)} } {\sin{\theta}} = \sqrt{2} = \dfrac {\frac{1}{2} \times \sqrt{2}} {\frac {1}{2}} = \dfrac {\frac{1}{\sqrt{2}} }{\frac {1}{2}} = \dfrac {\sin {45^\circ}} {\sin {30^\circ}} \quad \Rightarrow \theta = \boxed {30^\circ}$

Let BF be orthogonal to the side AC with point F on AC. As angle DCB equals 30, then from the right triangle we get $BF=BC \sin 30 ^ \circ = BC / 2 =BD=DF$ .

Then, $\angle FBD = 60 ^ \circ$ and $BF = FD$ , so $BFD$ is an equilateral triangle. This gives us $\angle FDB = 60 ^ \circ$ and $\angle FDA = 60 ^ \circ - 45 ^ \circ = 15 ^ \circ$ .

Since $\angle FAD$ and $\angle DFA$ are both $15 ^ \circ$ , so $FAD$ is an isosceles triangle with $AF = FD$ . Since $AD = FD = BF$ , thus $AFB$ is an isosceles triangle and $\angle FAB = 45 ^ \circ$ .

Thus, $\angle BAD = \angle BAF - \angle DAF = 45 ^ \circ - 15 ^ \circ = 30 ^ \circ$ .