Find missing value - geometry

Question

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.

Answer 1

If $\angle AEC = 2\theta$ then,
$$\angle AEF = 60^\circ + 2\theta \text { and } \angle EAC = 90^\circ - \theta$$

Now use the fact that $EA = EF$ and hence $\angle EAF = 60^\circ - \theta$

That leads to $\angle CAF = 30^\circ$ and finally $a = 15^\circ$

Answer 2

Since $EA=EC=EF$, a circle with center $E$ and radius $EA$ passes through $A, C, F$. Therefore $\angle CAF = 1/2 \angle CEF = 30^\circ$.

So $a= \angle CAD - \angle CAF = 45^\circ - 30^\circ = 15^\circ$.

Answer 3

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?

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