(1) First, it is proved that the triangle DCE is similar to the triangle EBF (there are 1 right angles respectively, and the angles corresponding to 1 group are complementary to the same angle [ef⊥de]).

So DC/EB = EC/EB (DC = AB = M, EC = X, EB = BC-EC = 8-X, BE = Y).

m/(8-x)=x/y

my=x(8-x)

y=(8x-x^2)/m(0<； x & lt8)(m & gt； 0)

(2) If m=8, then y = (8x-x 2)/8 = x-x 2/8 (0

It can be seen that the symmetry axis of this function is-b/2a =-1(-1/8 * 2) = 4.

Because a =- 1/8

Therefore, when X=4, the maximum value of y =4- 1/8*4*4=2.

(3) Because EF⊥DE, if △DEF is an isosceles triangle, it can only be DE=FE.

Because triangle DCE is similar to triangle EBF

DE=EF, so

Triangle DCE is equal to triangle EBF.

DC=EB，CE=BE

m=8-x，x=y

So m=8-x is enough (0