Current location - Training Enrollment Network - Mathematics courses - Mathematics for the senior high school entrance examination in Shanghai in 2009
Mathematics for the senior high school entrance examination in Shanghai in 2009
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.