A problem related to circle , altitude , triangle. Consider a $\triangle ABC.$ Draw circle $S$ such that it touches side $AB$ at $A$. This circle
passes through point $C$ and intersects segment $BC$ at $E.$
If Altitude $AD =\frac{21(\sqrt{3}−1)}{\sqrt{2}}\;$
and $\;\angle EAB = 15^\circ,$ find AC.
Here is a sketch that I made:

Here , we can calculate AC by Pythagorean theorem  if we find DC . This is where I'm stuck . How can I utilize the information that angle EAB is 15 deg ?
(This is not class-homework , I'm solving sample questions for a competitive exam )
 A: As user2345215 pointed out in a comment, a key point to this question is the fact that the circle touches $AB$ in $A$. So it isn't a mere intersection but instead a tangentiality. Even with this stronger constraint, the configuration isn't fully determined. If you fix $A$ and $D$, you can still move $B$ on the line $BD$. But $C$ isn't affected by changes to $B$'s position.

So given $A, D, B$ with a right angle at $D$, how can you construct the rest? You can use $\measuredangle EAB=15°$ to construct the line $AE$, and intersecting that with $BD$ you get $E$. Then you can construct the perpendicular bisector of $AE$ since any circle which passes through $A$ and $E$ has to have its center on that bisector. You can also construct a line through $A$ and orthogonal to $AB$. If the circle touches $AB$ in $A$, its center has to lie on that line. Intersecting these two lines gives you $F$, the center of the circle.
Now take a closer look. $GF$ is perpendicular to $AE$, and $AF$ is perpendicular to $AB$. Since $AB$ and $AE$ form an angle of $15°$, so do $AF$ and $GF$. So we have $\measuredangle AFG=15°$ and $\measuredangle AFE=30°$ Due to the inscribed angle theorem this tells you that $\measuredangle ACD=15°$ no matter where you place $B$. So you know that $AD=AC\cdot\sin15°$ which gives you
$$AC=\frac{AD}{\sin 15°}=42$$
A: Below is a scanned image of my solution worked out on paper.


