You are in luck. I answered this 24 hours ago, but not on this site.

The idea here is to first take FE and CE as lines but we do not assume that FE and CE are coincident (lie along the same line).

We then proceed to prove that the sum of angles AEF, AEB and BEC is equal to 180 degrees.

Because the equilateral triangles are attached to the sides of the square, every single straight line made out of solid lines in the diagram (AB, BC, CD, AC, AF, DF, AE and BE) are all equal in length.

- Triangle FDC is isosceles because FD = CD, so angle DFC = 15

- Angle AFD = 60, so angle AFC = 45 (we need this to find angle AEF)

- Angle BAE = 60, so angle DAE = 30 (we need this to find angle AEF)

- Considering triangle AEF, angle AEF = 180 - 45 - 60 - 30 = 45

- Angle AEB = 60 (and now we need to somehow find angle BEC to complete this proof)

- Angle ABE = 60, so angle EBC = 30

- Triangle EBC is isosceles because BE = BC, so angle BEC = 75

- Therefore, angle AEF + angle AEB + angle BEC = 45 + 60 + 75 = 180

so C, E and F lie on a straight line.