Geometry, triangles, and lengths

Question

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.

Answer 1

Hint:

1. Use the law of sines in $\triangle DBC$ to get a relationship between $\sin\angle BDC$ and $g$.
2. Use the law of sines in $\triangle DAC$ to get a relationship between $\sin\angle ADC$ and $a$ and $2c$.
3. Noting that $\sin\angle BDC=\sin\angle ADC$, use the previous two steps to get a relationship between just $a$ and $c$.
4. Use the Pythagorean theorem to get a second relationship between $a$ and $c$.

Answer 2

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, $AB =\sqrt{3+4}= \sqrt7$.

Answer 3

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 $A.$