This is mainly to understand that AF is equal to CG, and point O is the center of a square. So connecting BD, we can get that BF equals DG through ASA. Therefore, AF is equal to CG, let AF and CG be X, and circles O and FG intersect with another point P, so that PG is equal by CG folding, and the intersection point G of vertical line segments is AB and intersects with point R, FR is 8-2X, RG is 8, and FG is 4+2X.

Problem solving process

Solution: Make BC parallel lines through O, and cross AB to M and CD to N respectively.

AF=A'F，OF=A'F+r(OA ')，FM =AM-AF，OM=AM=4，r=2

According to Pythagorean theorem

At Rt△OMF, of 2 = FM 2+OM 2.

That is, (a 'f+r) 2 = (am-af) 2+om 2.

That is, (a 'f+2) 2 = (4-a 'f) 2+4 2.

The solution is A'F=7/3.

△OFM≌△OGN,∴OF=OG

∴a'g=fg-a'f=2fo-a'f=2r+a'f=4+7/3= 19/3