24.(1)解:3.(2)证明:'∠AEB+∠BED=180°,∠CFA十∠CFD=180°,且∠BED=∠CFD,·∠AEB=∠CFA,'∠BED=∠BAC,∴.∠ABE=∠BED-∠BAEI∠AEB=∠CFA,=∠BAC-∠BAE=∠CAF.在△ABE和△CAF中,∠ABE=∠CAF,AB =CA,.△ABE△CAF(AAS).(3n.1g15