Finding an angle between the side of a triangle and a segment from a point inside the triangle. Question given below:

ABC is a triangle and D is a point inside ABC such that:
$$
m(\widehat{DCB})=m(\widehat{CBD})=18^{\circ}\\
m(\widehat{ACD})=24^{\circ}\\
m(\widehat{DBA})=12^{\circ}\\
m(\widehat{DAC})=\alpha=?
$$
This is supposed to be a high-school level question. But i can't find the α. Is there something obvious i'm missing?
I checked that question is well-posed and α = 78 degrees. If i can just show that $|CD|=|CA|$ then i'm done. But that's the only step i can take in this question (If this can be called a step at all.).
 A: By the Trig Ceva Theorem,
$$\frac{\sin(12^\circ)}{\sin(18^\circ)}\cdot \frac{\sin(18^\circ)}{\sin(24^\circ)}\cdot\frac{\sin(\alpha)}{\sin(108^\circ-\alpha)}=1$$
hence:
$$\sin(96^\circ-\alpha)+\sin(120^\circ-\alpha)=2\cos(12^\circ)\sin(108^\circ-\alpha)=\sin(\alpha)$$
or:
$$2\cos(12^\circ)\sin(108^\circ)\cos(\alpha)=(2\cos(12^\circ)\cos(108^\circ)+1)\sin(\alpha)$$
or:
$$\tan(\alpha)=\frac{2\cos(12^\circ)\sin(108^\circ)}{2\cos(12^\circ)\cos(108^\circ)+1}$$
from which it is tedious (but not difficult) to check that $\alpha=78^\circ$ is the wanted solution. 
The identity:
$$\sin(18^\circ)+\sin(42^\circ)=\sin(78^\circ)=\cos(12^\circ)$$
is interesting in itself, but just follows from the Briggs formulas.
A: I solved this with elementary methods. In case of anyone interested comes and looking for answer and for archiving reasons, i post the solution:

Draw angle bisector of $\widehat{ACD}$ making $\overset{\Delta}{BEC}$ isosceles. Draw perpendicular from $E$ to $[BC]$ trisecting $\widehat{BEA}$ and making $\overset{\square}{ACDE}$ kite, therefore proving $|CA|=|CD|$.
