20 15-20 16 reasons for going to the eighth grade junior high school for math survey and test.

Solution: (1) Because of folding

So CB=CD=5, so OD= root sign (CD&; # 178; -OC & amp； # 178; )=3

According to simple proof, the triangle cod is similar to the triangle DAE.

Then by CO/OD=AD/AE? AE= 1.5？ So E(5, 1.5)

(2) the topic is easy to get, CR/QR=CB/BE=2.

(3) As can be seen from the question, tan Angle E=tan Angle CKE= 1/2 is easy to draw and analyze.

Then intersection c is CI parallel EK intersection CD at k point, and intersection I is IH vertical CD intersection CD at h point.

At this time, the tangent angle ICH = tangent angle CKE= 1/2.

So let IH be a, CH be 2a, and ci = (radical number 5) a.

And because there is a triangular IHD similar to triangular cod at this time.

So a=20/ 1 1 from IH/HD=OC/OD? So in the right triangle COI, OI= root sign (ci&; # 178; -OC & amp； # 178; )=8/ 1 1

Therefore, the resolution function of CI segment is y=- 1 1/2x+4.

Then the analytical formula of EK segment when k is the same in parallel is y=- 1 1/2+29.

And because the analytical formula of CD segment is y=-4/3x+4.

So column-11/2x+29 =-4/3x+4? The solution is x=6=m 6 = m 6 = m.