With sin∠F=3/5,

Let ce = 3x.

CF=5x，

∴ EF=4x，

ED=EC=3x，

∵FD^2=FC*FB，

∴(7x)^2=5x*(5x+24)，

x=5，

EF = 4x = 20

(2)AB and CD are not parallel.

Link BD,

△BDF∽△DCF。

∴BD/DC=DF/CF。

free

CF=25，

DF=35，

DC= 15 root number 2.

∴BD= 2 1 root number 2.

∵BD≠BC，

∴∠BDC≠∠BCD≠∠ABC，

∴ AB is not parallel to CD.

1, because sinF=3/5, then let CE=3x, then there are: CF=5x, EF=4x, ED=3x.

According to secant theorem: df 2 = cf * BF.

Then: (4x+3x) 2 = 5x (5x+24)

The solution is x=5.

Then, EF=5*4=20.

2. Obviously, △ECD is an isosceles right triangle.

∠ Equivalent cyclic density =45. If AB ∠ CD, then ∠ A = 45, then ∠ B = ∠ ACB = 67.5 = ∠ ECF.

∠F=22.5. Then, take a point G on e a so that ∠ EFG = 45, then CF is the bisector of ∠GFE. Therefore: GE=EF=20, EC= 15, GC=5, GF=20√2.

GF:GC=EF:EC

Substituting numbers is obviously not equal. So AB is not parallel to CD.

According to the universal formula: sinx = 2tg (x/2)/(1+TG (x/2) 2), cosx = (1-TG (x/2) 2)/(1+TG (x/2)).

Double angle formula:

sin2x=2sinx*cosx

cos2x=2cosx^2- 1= 1-2*sinx^2=cosx^2-sinx^2

tg2x=2tgx/( 1-tgx^2)

sin∠f=2tg(∠f/2)/( 1+tg(∠f/2)^2)=3/5

TG(≈F/2)= 3 (truncated value) or 1/3.

TG(≈F/2)= OB/BF = R/BF = 1/3，BF=3R

In the right triangle AEF, sin∠F=cos∠A=3/5.

According to the general formula, it can be obtained in the same way

TG(≈A/2)= 1/2 or-1/2 (truncated), sin(≈A/2)= 1/ radical number 5.

That is, TG(≈a/2)= ob/ba = r/ba = 1/2, and ba = 2r.

Sin (≈ A/2) = (BC/2)/AB =12/2r =1/radical number 5.

So R=6 times the root number 5.

△ECD is an isosceles right triangle, then OCED is a square. DE = R。

FE=FD+ED=FB+ED=3R+R=4R=24 radical number 5

AB and CD are not parallel.

△ECD is an isosceles right triangle, so ∠ECD=45 degrees, while ∠A is not equal to 45 degrees, so CD is not parallel to AB.