Finding the angle formed by the tangents of a triangle inscribed in a square

Question

In the following image, $ABCD$ is a square and $AEF$ is an isosceles triangle with $|AF|=|AE|$, such that $E$ and $F$ are on segments $BC$ and $DC$, respectively. Point $I$ is the barycentre of $AEF$ and circle $c$ has centre $I$ and is tangent to $FE$ in $J$. $AK$ and $AL$ are the tangents to circle $c$ from point $A$. Find the measure of $\angle{BAL}$.

Any ideas to this problem? I did notice that the points of tangency of lines $AK$ and $AL$ to circle $c$ form with point $J$ an equilateral triangle. I feel like that could be the key to the solution but I can't seem to find anything to go further (also it looks like those points form a parallel line to $FE$ but I can't find a succesful proof of that).

Answer 1

Because $I$ is the centroid, we have that $AI = 2IJ$. Letting $P$ and $Q$ be the $2$ tangency points, we then have that $\angle IPA = \angle IQA = 90^{\circ}$ and $AI = 2IP = 2IQ$. However, this means that $\triangle IPA$ and $\triangle IPQ$ are 30-60-90 triangles, so $\angle BAL = 45 - \angle IAP = \boxed{15^{\circ}}$

Answer 2

Hint: $AI=\frac 23AJ$. There is nothing to locate $J$ in the problem. If you trust the problem setter to ensure a unique solution, it cannot depend on where $J$ is. Probably the easiest place is $J=C$. Now figure out where $I$ is and the radius of the circle. If you are not willing to trust the problem setter, you need to demonstrate that the positions of $K$ and $L$ do not depend on where $J$ is.