Question 1199846
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{{{drawing(400,9200/27,-.2,2.5,-.2,2.1,
line(0,0,1,0), line(1,0,1+sqrt(3)/2,.5), line(1+sqrt(3)/2,.5,2.366025404,1.366025404),

line(1,0,2.366025404,1.366025404),
line(0,1,1,0),
line(1,0,1,1),line(1,1,1+sqrt(3)/2,.5),
line(2.366025404,1.366025404,3/2,1+sqrt(3)/2),
line(3/2,1+sqrt(3)/2,1,1), line(1,1,0,1),line(0,1,0,0),

locate(0,0,B), locate(1,0,C), locate(0,1.14,A), locate(.95,1.13,D),

locate(1.02,.92,60^o),
locate(1.47,2,G),locate(2.4,1.4,F),locate(1.9,.52,E) 
)}}}

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

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

Edwin</pre>