In isosceles △ABC, AB=AC=8, ∠ BAC = 120, and P is the midpoint of BC. Xiaohui holds a transparent triangular plate with an angle of 30, so that the vertex of the angle falls at point P and the triangular plate rotates around point P.

(1) As shown in figure 10( 1), it is proved that △BPE∩△CFP when two sides of a triangle intersect AB and AC at points E and F respectively.

(2) Operation: When the triangle rotates around point P to figure 10(2), the two sides of the triangle intersect with the extension line of BA and AC side at point E and point F respectively.

1) 1, are △BPE and △CFP similar? (Write conclusions only)

2) Explore 2. Are the connections EF, △BPE and △PFE similar? Try to explain why.

3) let EF=m and the area of △EPF be s, and try to represent s with an algebraic expression containing m. ..

( 1)

From AB=AC, ∠ BAC = 120.

∠b =∠c = 1/2( 180- 120)= 30。

By ∠ b+∠ EPB+∠ BEP =180 ∠ EPF+∠ EPF =180 ∠ EPF = 30 =∞.

∠BEP=∠CPF

∴△BEP∽△CPF

∴PF/PE=CP/BE

And ∵P is the midpoint of BC, that is, CP=BP.

∴PF/PE=BP/BE

That is, PF/BP=PE/BE.

At △BPE and △PFE, ∠EPF=∠B, PF/BP=PE/BE.

∴△BEP∽△PEF

(2) From (1), it can be known that △PFE∽△CPF.

EF/PF=PE/PC

That is, PE*PF=EF*PC.

s = 1/2 * PE * PF * sin 30 = 1/4 * PE * PF = EF * PC/4

EF=M, PC=AC* root number 3/2=4* root number 3

So S= root number 3*M