Let OP be perpendicular to AB and p, then AP=BP. AP-AG=BP-BH, that is, PG=PH, even PG and PH, it is easy to prove that triangle OPG is all equal to triangle OPH, and PG=PH is obtained. Angle PGO= angle PHO, angle DGB= angle FHA, so angle DGO= angle FHO. Then OM is perpendicular to CD in m and ON is perpendicular to EF in n, which proves that triangle OMG is equal to triangle ONH by (angle, angle, edge). Get OM=ON. So CD=EF (in the same circle or in the same circle, the chords with equal center distance are also equal)