Let PG=FE be perpendicular to AD and intersect with AD at G point, then GH is parallel to PB and intersects with AB at H point. Because PB is perpendicular to PE, PE is perpendicular to HG, and AH is perpendicular to PF, AHG is similar to FPE, and because AG=PG=HB, AH=AB-HB, PF=GD=AD-AG, AHG and FPE are congruent, so AG=FE, so PG =