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 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
$$
a = 4, \;\; b = 2\sqrt{3}, \;\; c = 2, \;\; d = \sqrt{3}, \;\; m = 1, \;\; n = 3.
$$
