(a)  Triangle CDE is equilateral triangle 
     (since it is isosceles CD = ED and the angle between 
      these congruent sides is 60°).

      Therefore, all its interior angles are 60° each.



(b)  angle ACE is 45° + 60° = 105°.

     angle CEF is 60° + 90° = 150°.



(c)  Triangle CEF is isosceles, since  CE = EF.

     Therefore, angle FCE =  =  = 15°.



(d)  < ACF = < ACE - < FCE = 105° - 15° = 90°.



ANSWER.  Angle ACF is 90° (right angle).

Since DC=DE and ∠CDE = 60o, △CDE is equilateral.
Therefore all the line segments except CF and AC are equal in length.

∠CED = 60o because △CDE is equilateral.
∠DEF = 90o because it is an internal angle of square DEFG
∠CED + ∠DEF = 60o + 90o = 150o
△CEF is isosceles because CE = EF.
The base angles of △CEF are equal in measure, and we can find them
by subtracting 180o-150o = 30o, and
then taking half and getting 15o.
So base ∠ECF = 15o,
∠DCE = 60o because △CDE is equilateral.
∠DCF = ∠DCE-∠ECF = 60o-15o = 45o
∠ACD = 45o because △ADC is an isosceles right triangle
∠ACF = ∠DCF + ∠ACD = 45o + 45o = 90o

Edwin