As shown in figure (1), in quadrilateral ABCD, AB = BC = CD, ∠ ABC = 78, ∠ BCD = 162. Let the extension lines of AD and BC intersect at point E, then ∠ AEB = _ _ _. (Shanghai in 2009

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.