Geometry, triangles, and lengths I'm a highschooler and this is the question:
"In a right triangle $\triangle ABC$, in which $\angle C = 90°$ and $\mid BC\mid < \mid AC \mid$, a line was constructed to go through point $C$ and crossing the hypotenuse in point $D$, $\mid AD\mid : \mid DB\mid = 2:1$. Given that $\mid BC\mid = \sqrt3$ and $\angle DCB = 30°$, calculate $\mid AB\mid$."
This is an image I have done to illustrate the question

I have tried to do it through many ways in the last few hours, like using the cosin theorem, using areas, I have even tried constructing right triangles inside the $\triangle ABC$ scaled 1:3 and 2:3 with the original one, but nothing worked for me. Every time I fail on calculating $\mid CD\mid$, and I might have done a calculational error. Can someone please help me? Sorry for any mistakes, this is my first post and English is not my first language. 
 A: Hint:


*

*Use the law of sines in $\triangle DBC$ to get a relationship between $\sin\angle BDC$ and $g$.

*Use the law of sines in $\triangle DAC$ to get a relationship between $\sin\angle ADC$ and $a$ and $2c$.

*Noting that $\sin\angle BDC=\sin\angle ADC$, use the previous two steps to get a relationship between just $a$ and $c$.

*Use the Pythagorean theorem to get a second relationship between $a$ and $c$.

A: Apply the sine rule to the triangles BCD and ACD respectively 
$$\frac{\sin 30 }{\sin \angle BDC }= \frac{BD}{BC}, \>\>\>\>\frac{\sin 60 }{\sin \angle BDC }= \frac{AD}{AC}$$
Take the ratio of the two equations to get
$$\frac{\sin 30 }{\sin 60 }= \frac{AC}{2\sqrt3} $$
which leads to $AC= 2$. Thus, $AB =\sqrt{3+4}= \sqrt7$.
A: There is an almost purely geometric solution.
I will not give the full answer but I think the following gives a big hint how to solve it.
In order to find the distance $|AB|$, we will construct the point $A.$
We know that $A$ is on the line through $C$ perpendicular to the line $BC.$
If we find the distance $|AC|$ then $|AB|$ can be found by the Pythagorean theorem.
The line through $C$ perpendicular to $BC$ is a locus of $A$.
There is a second locus of $A$ that consists of every point $P$ such that the line that passes through $C$ at a $30$-degree angle to $BC$ cuts the segment $BP$ in the ratio $1:2.$ Or in other words this locus consists of all points that are $3$ times as far from $B$ as $30$-degree line is, measured along some straight line through $B.$
If you can find that second locus, then find where it intersects the first locus (the perpendicular line), you have found $A.$
