The triangle ADE pair is folded into a triangle AFE, so AF=AD, angle AFG= angle AFE= angle ADE=90 degrees, and according to the square, angle B=90 degrees. In right triangle ABG and right triangle AFG, according to (H L) AG=AG, AB=AF, triangle ABG congruent triangles AFG.

According to the congruence of the above two triangles, let BG = FG = X. According to the known number of conditions, BC = CD = AB = 6, DE = EF = 2, CF = 4, CG = 6-X. In the right triangle CEG, according to Pythagorean theorem, the square of (6-X)+the square of 4 =(2+x).

Because FG=CG and angle GFC= angle GCF, according to the congruence of the first two triangles, angle AGB= angle AGF, and because angle AGB+ angle AGF is equal to angle GFC+ angle GCF and angle AGF= angle GFC, AG is parallel to CF.

Suppose the area of triangle FGC is equal to 3. Because CG=3, the height of the triangle must be 2, and because CE=4, according to the similarity, point F must be the midpoint of EG, which contradicts the previous FG=3 and EF=2.

To sum up: the right things one, two and three. Option c

Sorry, some symbols are not funny. They are arranged by junior high school mathematics. I am a teacher in a school in Wanzhou.