如图,BD平分∠ABC交AC于D,点E为CD上一点,且AD=DE,EF∥BC交BD于F.求证:AB=EF.
问题描述:
如图,BD平分∠ABC交AC于D,点E为CD上一点,且AD=DE,EF∥BC交BD于F.求证:AB=EF.
答
证明:作AM∥EF交BD的延长线于M,∵EF∥BC,∴BC∥AM,则∠M=∠DBC,∵BD平分∠ABC,∴∠ABD=∠CBD,∴∠M=∠ABD,∴AM=AB,∵AM∥EF,∴∠M=∠DFE,在△ADM和△EDF中∠ADM=∠EDF∠M=∠EFDAD=DE∴△ADM≌△EDF,...