A.D. ∨ B.C. ∠CEF=∠DAF

Because: AF is the bisector of ∠BAD, and there is ∠FAB=∠DAF, so ∠CDA=∠CEF.

That is, the triangle CEF is an isosceles triangle, where ∠CDA=∠CEF, CE = CF

Question 2: After proving that triangle DGC and triangle Berg are similar triangles, there are: ∠CGD=∠EGB, DG = BG and ∠ CGA = 90, so ∠ BGD = 90, so triangle BGD is an isosceles right triangle ∠ BGD = 90, and ∠. The proof that triangle DGC and triangle BEG are similar triangles is as follows: AF is the bisector of ∠BAD, ∠ BAC = 90, from which ①AB=BE=DC and ∠ DCB = 90 are derived. According to the question 1, it is proved that the triangle ECF is an isosceles right triangle, G is the midpoint of the hypotenuse EF, and ② EG = GF. ③ ∠ beg =180-∠ cef =135 = ∠ BCD+ECD = ∠ DCG. That is, a triangle with two equal sides and included angles is similar triangles.

Question 3: Extend the intersection of AB and FG to point H, the parallelogram AHFD is a diamond, HD=DF, BH = CF = FG, ∠ CFG = ∠ BHD = 60, and the triangle BHD is similar to the triangle GFD, ∠ BDH = ∠ GDF, ∠BDG =∞.