$\angle{B}=45^{\circ}$ in $\triangle{ABC}$, $D$ is on bisector of $\angle{A}$ and $AD=BC, \angle{ACD}=30^{\circ}$. Find$\angle{A}$. 
background: chatting with people in mathematics problems group, and mentioning some geometry problems that no pure geometric approach being found, and someone gave this problem as an example.
motivation: To find a solution of this hard problem.
relevant definitions: Angle bisector, i bet all you know what it is.
source: No to my knowledge. Someone gave to me on social media group.
possible strategies: Need find a good construction. I can see extending AD and CD to meet the sides of the triangle, actually there are four points cyclic.

your current progress: I can get trigonometric equality from Ceva's theorem, and from making diagram it is easy to see the answer is 30 degree.
$sin(2x)sin(x+30^{\circ})=\dfrac{\sqrt{2}}{4} $
why the question is interesting or important: because it is hard to construct a pure geometric approach to solve it.
etc: i cannot think of any etc yet.
Is there a pure geometric approach to solve this problem? Thanks.
 A: 
In the figure, if we reflect $D$ about $AC$ to get $Y$ as shown, we have $AY=BC$. Thus it suffices to prove that $\alpha=15^o$ when $AY=BC$, $\angle YCA=30^o$ and $\angle ABC=45^o$
Note that $\alpha \lt 60^o \implies AC \gt AY$ and $AY=BC \implies AC \gt BC \implies \angle CAB \lt 45^o$

In the second figure, it is obvious that $AE=CX$
The proof is based on the method of exhaustion. That is we prove that $\alpha \gt 15^o$ and $\alpha \lt 15^o$ are both impossible and hence $\alpha$ must be $15^o$
Assume that $\alpha \gt 15^o$.
We combine the $2$ figures as follows:

Note that $\angle EAY=\alpha-15^o$, $\angle BCX=2\alpha-30^o$
Also note that $A, B, X, C$ are concyclic and hence $\angle XCB=2\alpha-30^o$
Let the angle bisectors of $\angle XAB$ and $\angle XCB$ meet at $W$ and $CW$ cuts $AB$ at $Z$
Note that $A, B, W, X, C$ are concyclic.
Since  $\angle CAB $ is acute and $A, B, W, C$ are concyclic therefore $\angle CWB = 180^o -\angle CAB $ is obtuse.
Hence in $\Delta CWB$, $\angle CWB \gt \angle CBW$.
Consequently $\color{red}{BC \gt CW}$
$\Delta CXZ \cong \Delta AEY \implies AY=CZ$
Since we are given that $AY=BC$, therefore $BC=CZ$
But obviously $CZ \lt CW$
Thus  $\color{red}{BC \lt CW} $
The two statements colored red obviously contradict each other.
Hence $\alpha \gt 15^o$ is imposssible.
By similar arguments, $\alpha \lt 15^o$ is impossible.
Hence $\alpha = 15^o$
Job is done.
