证明:∵DF⊥AC,BE⊥AC,∴∠DFA＝∠BEC＝90° ∵AD∥BC,∴∠DAF＝∠BCE,AD＝BC,∴△DFA≌△BEC,∴∠ADF＝∠CBE,∵HF＝½AD＝½BC＝EG,又∠HFD＝∠HDF＝∠GBE＝∠GEB,∴∠HFD＋∠DFE＝∠GEB＋∠BEF,即∠HFE＝∠GEF,∴HF∥EG,∴四边形EGFH为平行四边形