Are you smarter than me? 54

Let $\angle BAD$ and $\angle BAC$ be $\theta$ and $\alpha$ respectively. Then $\alpha = \theta + 15^\circ$ . Using Sine Rule, we have:

$\dfrac {\sin{\alpha}}{a} = \dfrac {\sin{30^\circ}}{c}\quad \Rightarrow \sin{\alpha} = \dfrac {a}{2c}$

$\dfrac {\sin{\theta}}{\frac {a}{2}} = \dfrac {\sin{45^\circ}}{c}\quad \Rightarrow \sin{\theta} = \dfrac {a}{2\sqrt{2}c}$

$\Rightarrow \sin {\theta} = \dfrac {\sin{\alpha}}{\sqrt{2}}\quad \Rightarrow \sqrt{2} \sin {\theta} = \sin {\alpha}$

$\Rightarrow \sqrt{2} \sin {\theta} = \sin {\theta+15^\circ} = \sin {\theta} \cos {15^\circ} + \sin {15^\circ}\cos {\theta}$

$\Rightarrow \tan {\theta} = \dfrac {\sin {15^\circ}}{\sqrt{2}-\cos{15^\circ}} = \dfrac {1}{\sqrt{3}} \quad \Rightarrow \angle BAD = \theta = 30^\circ$ or $120^\circ$ .

But if $\theta = 120^\circ$ , $BD > CD$ which is unacceptable. Therefore, $\theta = \boxed{30^\circ}$

$Let~ x=\angle DAB, ~~\angle ABD=180 - x - 45 .~~\angle CAD =45 - 30=15.\\ Applying~ Sin~ Law, ~ in~ \Delta s~ ABD~and~ACD, \\ \dfrac{AD}{\frac {BC} 2}=\dfrac{Sin(180-x-45)}{Sin(x)}=\dfrac{Sin(30)}{Sin(15)}\\ \dfrac{1 - cot(x)}{\sqrt2} =\dfrac{Sin(30)}{Sin(15)}\\ Solving~ we~get~\angle BAD= \Large~~~~~~~~ \color{#D61F06}{30^o}$

You Can Simply use m-n Theorem In ABC. Let BAD = x And Easily we can see that ADC = 135 and CAD = 15 Apply m-n theorem (1+1)cot 135 = 1 cotx-1 cot15 2cot135=cotx-cot15 cotx = 2cot135+cot15 =root 3 x = 30 because x greater than 180 is not acceptable