(1) find ∠ ECB = 15, ∠ DCF = 60, DF=3√3, DC=6, deduce AB=DF=3√3, BC=3√3, and find AD = DF = 3 √ 3.

(2) After crossing point C, make the extension line of CM perpendicular to AD at point M, and then extend DM to point N. Prove that MN=BE and then △ DEC △ DNC, and get ED=EN, then the answer can be deduced.

explain

Solution:

( 1)

∠∠BEC = 75 degrees, ∠ABC=90 degrees.

∴∠ECB= 15

∫∠ Equivalent circulating density =45

∴∠DCF=60

In Rt△DFC:

∠DCF=60，FC=3

∴DF=3√3,DC=6

Judging from the title:

The quadrilateral ABFD is a rectangle.

∴AB=DF=3√3

AB = BC

∴BC=3√3

∴BF=BC-FC=3√3-3

∴AD=DF=3√3-3

∴C trapezoidal ABCD=3√3×2+6+3√3-3=9√3+3.

Answer: The circumference of trapezoidal ABCD is 9√3+3.

(2)

Prove:

Extend EB to G, make BG=CF, and connect CG.

∠∠CBG =∠DFC = 90，BC=FD

∴△BCG≌△FDC

∴∠ 1=∠2

∠∠2+∠DCF = 90

∴∠ 1+∠DCF=90

∠∠DCE = 45

∴∠ECG=45

∴∠DCE=∠ECG

∴△DEC≌△EGC

∴ED=EG

∴ED=BE+FC