(1) Proof: According to the meaning of the question ∠ CDF = 180-∠ BDE-∠ EDF.

= 180 -∠BDE-∠B

= < bed

Because AB=AC, ∠ b = ∠ C.

So △BDE∽△CFD

(2) Proof: Since DF=EF, ∠ EDF = ∠ FED = ∠ B = ∠ C.

So △ Fed ∽△ABC

Because BD=CD

So FD/AB=ED/BC=ED/2BD.

ED/FD=BD/FC because △BDE∽△CFD.

So FD/AB=ED/2BD=FD/2FC.

So AB=2FC=AC.

So AF=FC

Because BD=CD

So DF is the center line of △ABC, so DF//AB

(3) solution: ∠ BAC = 180-∠ AFD because DF//AB.

Because de⊥ac∠AFD+∠EDF = 90.

Because EDF = b = C.

So ∠ EDF = ( 180-∠ BAC)/2。

So ∠ BAC =180-[90-(180-∠ BAC)/2] =180-∠ BAC/2.

So ∠ BAC = 120.

(4) Solution: ① In the isosceles triangle ABC, AB=AC=5, BC=8.

So ∠B=∠C=arcsin(3/5)

So ∠EDF=∠B=arcsin(3/5)

Because △BDE∽△CFD, DE/DF=BE/DC=x/4.

So DF=4DE/x

△ bed, according to cosine theorem

DE^2=BE^2+BD^2-2*BE*BD*cos∠B

=x^2+ 16-32x/5

So s △ def = y = (1/2) * de * df * sin ∠ EDF.

=( 1/2)*(4/x)*DE^2*(3/5)

=(6/5x)*(x^2-32x/5+ 16)

= 6x/5- 192/25+96/5x(x & gt； 0)

②△AEF is an isosceles triangle.

The first case: AE=EF

Because AE=BE-AB=x-5,

And ef 2 = de 2+df 2-2 * de * df * cos ∠ EDF.

=x^2+ 16-32x/5+( 16/x^2)*(x^2+ 16-32x/5)-(8/x)*(4/5)*(x^2+ 16-32x/5)

=(x^2+ 16-32x/5)( 1-32/5x+ 16/x^2)

EF=x+ 16/x-32/5

So x-5=x+ 16/x-32/5.

x=80/7

The second case: AE=AF

Because △BDE∽△CFD, BD/FC=BE/DC.

So FC= 16/x

So af = AC-FC = 5-16/x.

So x-5=5- 16/x

x^2- 10x+ 16=0

X=2 (truncated) or 8.

The third case: AF=EF

5- 16/x=x+ 16/x-32/5

x+32/x-57/5=0

5x^2-57x+ 160=0

(x-5)(5x-32)=0

X=5 (truncated) or 32

To sum up, x=80/7 or 8 or 32.