point out

The intersection f makes FM parallel to AC and intersects BC at m.

Then it is proved that the triangular FMD is similar to the triangular ECD.

Method 2; Connecting AD, because CD=BC, it is easy to know that C is the midpoint of BD, and because F is the midpoint of AB, AC and DF intersect with E, so E is the center of gravity of triangle ABD, which can be easily found by using the nature of the center of gravity.

Method 3; Connect AD and FC, and solve the problem by using the property of neutral line.

The final result of the above three methods is 2 to 3.