As shown in Figure 1, in the rectangular trapezoid ABCD, ad ‖ BC, ab ⊥ BC, ∠ DCB = 75 ° and the other vertex e of equilateral △ DCE with CD as one side is on the waist ab (1) (2) AB = BC; (3) as shown in Figure 2, if f is a point on the line CD, ∠ FBC = 30 °, the value of dffc can be obtained

As shown in Figure 1, in the rectangular trapezoid ABCD, ad ‖ BC, ab ⊥ BC, ∠ DCB = 75 ° and the other vertex e of equilateral △ DCE with CD as one side is on the waist ab (1) (2) AB = BC; (3) as shown in Figure 2, if f is a point on the line CD, ∠ FBC = 30 °, the value of dffc can be obtained

(1) From the equilateral ⊥ BC and ad ∥ BC, we can get ∵ DAB = 90 ° and ∥ AED = 45 ° (2) it is proved that from (1) we know ∵ AED = 45 ° and ∥ ad = AE, so point a is on the vertical bisector of line de. from ⊥ DCE is an equilateral triangle, we can get CD = CE, so point C is also on the vertical bisector of line De It is the vertical bisector of segment De, that is, AC ⊥ de. connecting AC, ∵ AED = 45 °, and ∵ ab ⊥ BC, ∵ ACB = 45 ° and ∵ Ba = BC. (3) ∵ FBC = 30 ° and ∵ ABF = 60 °. The extension line connecting AF, BF and ad intersects at point G,