In rectangular trapezoid ABCD, ad is parallel to BC, AB is vertical to BC, and the angle DCB is 75 degrees. The other vertex e of equilateral triangle with CD as one side is on the waist ab Find AB = BC If f is a point on the line CD, the angle FBC = 30 ° and DF = FC are proved

In rectangular trapezoid ABCD, ad is parallel to BC, AB is vertical to BC, and the angle DCB is 75 degrees. The other vertex e of equilateral triangle with CD as one side is on the waist ab Find AB = BC If f is a point on the line CD, the angle FBC = 30 ° and DF = FC are proved

Ding Sucheng, DSC, Hello: solution: (1) ∵ BCD = 75, ad ∥ BC ∥ ADC = 105, from the equilateral △ DCE, we can know: ∵ CDE = 60, so ∥ ade = 45, from ab ⊥ BC, ad ∥ BC, we can get: ∵ DAB = 90, ∵ AED = 45 ∥ ad = AE, so point a is on the vertical bisector of line de. from the equilateral triangle of △ DCE, we can get: CD = CE, so