Answer: 2 1
Solution (not in the original counseling materials, here is the solution process I did):
As shown in Figure (2), BC and AB respectively intersect at point F through parallel lines of point A and point C to connect DF.
∫∠ABC = 78,ab∨cf ┃∴cd=cf=af
∴∠ BCF = 102∴∴△ CDF is a regular triangle.
∫≈BCD = 162 ┃∴af=df,∠cfd=∠cdf=60
∴∠DCF=60,∠DCE = 18 ┃∴∠adf=﹙ 180 -∠afd﹚÷2=2 1
∫ab∨cf,BC∨af,AB=BC ┃∴∠ADC=39
∴ quadrilateral ABCF is a diamond, ∠ AFC = 78┃∵∠ ADC is the outer corner of △DCE, ∠ DCE = 18.
∴BC=CF=AF ┃ ∴∠AEB=2 1
bc = cd
I can't send this picture yet. Give me a website. This is it:/%C6% AE% D7% DF% B5% C4% C1%F8% D0% F5/Album/Item/022AA3669DAE 790E39DD.html # img = 0222AA3669DAE 799.
This is a competition problem in 2009. I hope the O(∩_∩)O~ dug up in my tutoring materials will help.