Consider $\triangle ABC$ with a point $D \in BC$. If $\triangle ABD \sim \triangle ADC$ in ratio $\frac {1}{\sqrt3}$. Find the angles of $ABC$
My try
Well, the first thing i saw is if $ABD$ and $ADC$ are similar, then if i'm not wrong, $AD$ is the angle bisector of $ABC$, because $\angle$DAC =$\angle$BAD so i played with the similarity ratio and put $x$ and $y$ to some lengths.
My Measurements
Ex:$\vartriangle$$ABD$ $\sim$ $\vartriangle$$ADC$
then $$\frac{AB}{AD}=\frac{BD}{DC}=\frac{AD}{AC}=\frac{1}{\sqrt3}$$
Let $AB=x$
$$\frac{x}{AD}=\frac{1}{\sqrt3}$$
$$x\sqrt3=AD$$
I did this with all the sides of the triangles, then i applied Angle bisector theorem with the measures and i ended with $\sqrt 3=3$ (obviously a wrong equality).
Sorry for not putting all my calculations here, but i think it doesn't matter because they are wrong.
I think my mistake is in the use of the ratios. Are my approaches good and i just put bad the ratios, or i missed something? any hints?
No trigonometry allowed
We should notice that there is a unique triangle that satisfies these conditions. For example if we try to take $\angle ABC > 90^\circ$, then we cannot have $\triangle ABD \sim \triangle ADC$ because $\angle ADC = \angle ABD + \angle BAD$.
If we try to take $\triangle ABC$ acute, then if $AD$ is not altitude of $\triangle ABC$, then either $\angle ADC > 90^\circ$ or $\angle ADB > 90^\circ$. For both cases, one of $\triangle ABD$ and $\triangle ADC$ is acute while the other one is wide-angle triangle. If $AD$ is altitude, for acute $\triangle ABC$, we can satisfy $\triangle ABD \sim \triangle ADC$ with $\angle BAD = \angle DAC$ but clearly, this gives us the ratio $1$, which is not we want.
Therefore, the triangle we seek is a special one where $\angle BAC = 90^\circ$. After that, if we place the angles as the following in order to have a ratio different than $1$, we can see that $\triangle ABC$ is $30^\circ-60^\circ-90^\circ$ triangle (WLOG, we can take $|AD| = 1$ and find the corresponding lengths with respect to $|AD|$).