It can be proved that △ FCE △ EBG (angular edge), that is △ CEF = △ BEG, then they are diagonal. We can see that FEG is a straight line. In △AFG, △ AFE △ age can be proved. Which means AF=AG. For a square, AB=BC, then AF = AB+BG = BC+FC. Over.

2. Answer: directly answer the second question, ∠1= ∠ DBE+∠ BDC = ∠ DBA+∠ Abe+∠ BDC.

From the 1 th answer, we know △ Abe △ ADC, that is, ∠ Abe = ∠ ADC.

Then ∠1= ∠ DBA+∠ ADC+∠ BDC = ∠ DBA+∠ BDA = 60+60 =120 (because △ABD is equilateral △).

3. Answer: I don't know if your diagram is misleading, 1, I think BH=CK.

(2) The area of quadrilateral CHDK should change from small to large, and then decrease.