As shown in the figure, there is an isosceles triangle cake (AB=BC, and BC≠AC) with chocolate evenly distributed on the edge of the cake, and Xiaoming and Xiaohua decide.

(1) According to the three-line unity of isosceles triangle, it is enough to make the median vertical line BD of line segment AC.

(2) Xiaomao will not succeed. The straight CD may divide the area of △ABC equally. If the circumference is also divided equally, AC=BC, which conflicts with AC≠BC in the question, so it will not succeed.

(3)① If the straight line passes through the vertex, the perpendicular line on the AC side is the demand.

② If the straight line does not cross the vertex, it can be considered in the following three cases: (a) The straight line intersects BC and AC at E and F, CF=5 and CE = 3 respectively; (b) A straight line intersects with AB and AC at m, n, AM=3, AN=5, and (c) A straight line intersects with AB and BC at p and q respectively, which does not exist. Then there are three eligible straight lines * * *. Solution: (1) Just make it the middle vertical line BD of the line segment AC.

(2) It will not succeed.

If the straight line CD bisects the area of △ABC, then S △ ADC = S △ DBC.

As shown in Figure 2, the crossing point C is CE⊥AB and the vertical foot is E. 。

Then 12AD? CE=

12BD? CE

Then BD=AD,

∵AC≠BC，

∴AD+AC≠BD+BC

It won't succeed.

(3) Classification discussion:

① As shown in Figure 3, if the divided straight line passes through the vertex of △ABC, it can be known from (1)(2) that only the straight line passing through the vertex of the isosceles triangle can be an "equal product perimeter".

That is, the middle vertical line BD on the side of bottom AC is the "equal division product perimeter" of △ABC at this time.

② If the dividing line does not pass through the vertex of △ABC, the dividing line divides ABC into a triangle and a quadrilateral. There are three situations:

(a) the straight line EF intersects BC and AC at e, f,

If the straight line EF bisects the perimeter of the triangle 16, the sum of CF and CE is 8.

Let CF=x, then ce = 8-x.cb = 5, CG=3, BG=52-32=4,

∵EH∥BG，

∴△CEH∽△CBG，

∴EHBG=ECBC，

∴EH4=8-x5，

EH=45(8-x)，

If the areas of the two divided parts are equal, then

S△CEF=6，

That is 12? x?

45(8-x)=6，