证明:$(2)$∵$AF $平分$∠ BAC,$$DE$平分$∠ BDC,$
∴$∠ BAC=2∠ BAF=2∠ CAF,$$∠ BDC=2∠ BDE=2∠ CDE。$
$ $由$(1)$得$∠ BDC=∠ BAC+∠ B+∠ C,$
又∵$∠ B=∠ C,$
∴$2∠ CDE=2∠ CAF+2∠ C,$
∴$∠ CDE=∠ CAF+∠ C。$
∵$∠ CGF=∠ CAF+∠ C,$
∴$∠ CDE=∠ CGF,$
∴$AF// DE。$