Find angles in a triangle, with two similar triangles with scale factor $\sqrt{3}$ 
Triangle ABC has point D on BC which creates triangle ABD and ACD. They differ with the scale factor $\sqrt{3}$. What are the angles?

I know ADB and ADC cannot be right, as it shares the side AD and cannot have a scale factor of $\sqrt{3}$.
I have tried to approach it with $\frac{AD}{DC}=\frac{BD}{AD}=\sqrt{3}$ (or $\frac{AD}{AC}=\frac{BA}{AD}=\sqrt{3}$). Meaning that $DC=x$ $AD=\sqrt{3}x$ $BD=3x$. This is where I get stuck. Because I dont know if the triangle is right angled, and cannot use Pythagoras. I dont have any angles, hence cannot use $\frac{sin(a)}{angleA}$. How do I solve this?
 A: *

*Let $\Delta ABD\sim\Delta ACD$. 


Hence, $\frac{AB}{AC}=\frac{AD}{AD}$, which is impossible.


*Let $\Delta ABD\sim\Delta ADC$. 


Hence, $\measuredangle ABD=\measuredangle ADC,$ which is impossible.


*Let $\Delta ABD\sim\Delta CAD$. 


Hence, $\measuredangle ADB=\measuredangle CDA=90^{\circ}$ and since $\frac{AD}{CD}=\sqrt3$, we have $\measuredangle C=60^{\circ}$
and from here $\measuredangle B=30^{\circ}$ and $\measuredangle BAC=90^{\circ}$.


*Let $\Delta ABD\sim\Delta CDA$.


Hence, $\measuredangle ABD=\measuredangle CDA$, which is impossible.


*Let $\Delta ABD\sim\Delta DAC$.


Hence, $\measuredangle BAD=\measuredangle ADC$, which says $AB||BC$, which is impossible.


*Let $\Delta ABD\sim\Delta DCA$, which gives $AB||BC$ again. 


Done!
A: $ABD$ and $ACD$ are similar and differ by scale factor $\sqrt3$.
We therefore know that $\angle ADB=\angle ACD\lor\angle ADB=\angle CAD\lor\angle ADB=\angle ADC$.
Let us consider the first case, where $\angle ADB=\angle ACD$.
Then $AD \,||\, AC$. But this is impossible unless $C$ and $D$ coincide which is also impossible due to the scale factor.
Let us consider the second case, where $\angle ADB=\angle CAD$.
Then $\angle ABD\neq\angle BAD$ as ${AD\over AD}\neq\sqrt3$. Therefore $\angle ABD=\angle ADC$ leading to a similar contradiction as before.
Hence we conclude $\angle ADB=\angle ADC$.
$B,C,D$ are collinear $\implies\angle ADB=\angle ADC = \frac{\pi}{2}$.
$\angle BAD\neq\angle CAD$ as ${AD\over AD}\neq\sqrt3$. Therefore $\angle BAD=\angle ACD$ and $\angle ABD=\angle CAD$.
wlog, let $ABD$ be the larger triangle and let $CD = x$.

You should now be able to fill out all of the lengths using the ratio of lengths of the similar triangles and Pythagoras (also, notice $\angle BAC = {\pi\over2}$).
You'll end up seeing you have $1 : \sqrt3 : 2$ right-angled triangles which you should know the angles of.

Finally, you may want to convince yourself that such a construction is possible. The way the question is set up implies that there is a solution, so there is no need to check once you find the only possibility.
However, outside of such a context, this may simply be done by first constructing $ACD$ and then letting $B$ be a point collinear to $CD$ such that $\angle BAC ={\pi\over2}$. Next, show that the triangles are similar and have a scale factor $\sqrt3$.
A: A solution is given by dropping an altitude to the hypotenuse of a 30-60-90 right triangle.  Is this the only one?  To find out we define the six quantities $a = BC$, $b = CA$, $c = AB$, $d = AD$, $m = BD$, and $n = DC$.  There are four ways in which the triangles $ABD$ and $ACD$ could be similar:
\begin{align}
1)& \;\; \frac{b}{c} = \frac{d}{m} = \frac{n}{d} = \sqrt{3},\\
2)& \;\; \frac{b}{m} = \frac{d}{c} = \frac{n}{d} = \sqrt{3},\\
3)& \;\; \frac{b}{d} = \frac{d}{c} = \frac{n}{m} = \sqrt{3},\\
4)& \;\; \frac{b}{d} = \frac{d}{m} = \frac{n}{c} = \sqrt{3}.\\
\end{align}
(The other two permutations have $d/d = \sqrt{3}$, so we can rule them out immediately.)  We need to solve for six quantities.  Each of the cases above reduces this to three.  The relationship $a = m+n$ reduces it two.  Setting one of the lengths to 1 reduces it one.  So we only need one more relationship.  For this we use Stewart's Theorem:
$$
man + dad = bmb + cnc.
$$
Cases 1 through 3 each have 1 solution with positive side lengths, and case 4 has 2 solutions.  However, only in case 1 do the three resulting triangles all satisfy the triangle inequality.  This solution is the 30-60-90 right triangle
$$
a = 4, \;\; b = 2\sqrt{3}, \;\; c = 2, \;\; d = \sqrt{3}, \;\; m = 1, \;\; n = 3.
$$
A: Hint:
$ABC$ is a rectangular triangle with $\hat A=90°$ and $\hat B=60°$ . 
Take  $D$ such that $AD$ is the height of the triangle with respect to $BC$.
