Note: As you said, BE and CD intersect at G point (obviously both B point and G point are on BE); BE and AF intersect at point g' (obviously point g' is on BE);

Assuming that points B, G and G' are all on the BE, it is obvious that the straight line * * *! Do you need to prove it again? !

The ideas and methods are wrong.

What should be clear is the three-point line of point A, point G and point F..

Prove: let vector AB= vector a, vector AC= vector b,

Vector BE= vector AE- vector AB= 1/2* vector AC- vector AB= 1/2* vector b- vector A,

Vector CD= vector AD- vector AC= 1/2* vector AB- vector AC= 1/2* vector a- vector B. . . . . . . . . . (*),

Because b, g and e are three-point lines, vector BG=m* vector BE=m/2* vector b-m* vector a(m is a real number),

Vector DG= vector DB+ vector BG= 1/2* vector AB+ vector BG= 1/2* vector a+(m/2* vector b-m* vector A).

=( 1/2-m)* vector a+m/2* vector b. . . . . . . . . . . . . . . . . . . . . . . . . (**),

Because the three-point * * line of C, G, D, G and D consists of (*) (* *), (1/2) * (m = 2/3 2)-(-1) * (1/2-).

So vector BG=(-2/3)* vector a+ 1/3* vector B,

Vector AG= vector AB+ vector BG= vector a+[(-2/3)* vector a+ 1/3* vector b]= 1/3* (vector a+ vector b),

Vector AF= vector AB+ 1/2* vector BC= vector AB+ 1/2* (vector AC- vector AB).

= 1/2* (vector AB+ vector AC)= 1/2* (vector a+ vector b)=3/2* vector AG.

So the three-point line of a, g and f,

So the three midlines of the triangle intersect at one point.