The Hanging Angle

I solved by sine rule; later I will edit this post and write the full solution. (Need to sleep now).

Applying sine rule in both triangles ABD and ADC, I got:

$\frac { \sin { x } }{ \sin { 45 } } =\frac { \sin { (x+45) } }{ 2\sin { 30 } }$

Then, it´s not hard to get:

$\cos { 15 } +\sin { 15 } \cdot \cot { x } =\sqrt { 2 }$

and doing a bit more math, we get

$\cot { x } =\sqrt { 3 }$ .

So, $\angle BAD = \boxed{30°}$

We draw $BE\perp AC$ $(E \in AC)$ .

Because $\angle ADB=\angle DAC +\angle DCA$ , we then have $\angle DAC=15^o$

Because $BE\perp AC$ and $\angle ACB=30^o$ , we then have $\angle EBD=60^o$ and $BE=\frac{BC}{2}=BD$ , therefore we have:

• $\angle EDA=15^o$ because $\angle EDB=60^o$ and $\angle ADB=45^o$
• $ED=BD$

Triangle $AED$ has $\angle DAE=\angle EDA = 15^o$ then $EA=ED$ therefore $EA=EB$ so $\angle BAE$ must be $45^o$ . Finally, we have $\angle BAD=30^o$ because $\angle BAD+\angle DAE=45^o$