∵ A circle with point F as the center and line segment FA as the radius is tangent to BC.

∴AF=EF

FC & ampsup2 Council of Europe. sup2+EF & amp； sup2

∴fc&； Sup2 Council of Europe. sup2+AF & amp； sup2？ ( 1)

AC = 3？

∴AF=AC-FC=3-FC？ (2)

∫∠B =∠C (isosceles triangle with equal base angle)?

∴cos∠C=cos∠B= 1/3

∫∠FEC = 90

∴EC=FC/3？ (3)

Substituting (2) and (3) into (1) gives:

FC & ampsup2=(FC/3)& amp； sup2+(3-FC)& amp； sup2

FC & ampsup2= FC & ampsup2( 1/9)+? (3-FC)& amp； sup2

(8/9)FC & amp； sup2=(3-FC)& amp； sup2

Both sides are fc &；; sup2(FC≠0)

8/9=(3/FC- 1)sup2(4)

∵FC & lt； three

∴3/fc>； 1

∴3/fc- 1>； 0

Both sides of (4) can be obtained at the same time.

∴3/FC- 1=(2/3)√2

∴3/FC=(2√2+3)/3

∴FC=9/(3+2√2)=9(3-2√2)

∴EC=FC/3=3(3-2√2)

AB = AC = 3？ cos∠B= 1/3

∴bc/2=ab×cos∠b=3*( 1/3)= 1

∴BC=2

∴BE=2-EC=2-3(3-2√2)=2-9+6√2)=？ 6√2-7

∫cos∠B = 1/3

∴BD=BEcos∠B

=( 1/3)[? 6√2-7]

=2√2-7/3

So when BD=2√2-7/3, the circle with point F as the center and line segment FA as the radius is tangent to BC.

(2) As shown in Figure 2.

∫x = BD

∴BE=x/cos∠B=？ x/( 1/3)=3x

FC=EC/cos∠C=？ EC/cos∠B=3EC

EF & Sup2 = FC & Sup2-EC and Council of Europe; sup2

=(3EC)& amp； Sup2- EC and Council of Europe; sup2

= 8EC & ampsup2

= 8(2-BE)& amp； sup2

= 8(2-3x)& amp； sup2？ (2 & gt3x)

GEC =∠b+∠ bromodiphenyl ether

∴∠GEF+90 =∠B+90

∴∠GEF=∠B

You can get ∠GFE=∠C in the same way.

∠∠B =∠C

∴∠GEF=∠GFE

△EFG is an isosceles triangle

∫cos∠B = 1/3

∴sin∠b=√[(3&； sup2- 1 & amp； sup2)/3 & amp； sup2]=( 1/3)√[(3 & amp； sup2- 1 & amp； sup2)=( 1/3)√8

∴tan∠B=√8=2√2

△ The bottom of △EFG is EF.

Let the height be h,

Then h = (ef/2) * tan ∠ GEF = (ef/2) * 2 √ 2 = ef √ 2.

△ the area of △EFG is y.

y = EF * h/2 = EF & amp； sup2√2/2

= 8(2-3x)& amp； sup2√2/2=4√2? (3x-2) and sup2

The analytical formula of triangle area is

Y=4√2？ (3x-2) and sup2

Domain:

2 & gt3x

∴x