∵ 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