Find missing value - geometry We are given square $ABCD$ and equilateral triangle $CEF$.
We are looking for angle $a$.

I have tried everything but no clue.
All I managed to find is that $EC=EA$ because $AC$ and $DB$ are the square's diagonals, which are perpendicular and bisect each other, so $EB$ is perpendicular bisector of $AC$, so all points on it are equidistant from $A$ and $C$. So triangle $EAC$ is isosceles.
Geogebra says that the angle in question is constant and $a = 15^\circ$.
So we need to prove that angle $FAC = 30^\circ$.
Any ideas?
Thank you very much.

Hint: Rotate both the square and the triangle clockwise by 60° around $C$. What is the image of $E$? What line does it belong to?

Edit. I am going to provide the image. It should be clear what to do now.

