如图,已知线段AB、CD相交于点O,AD、CB的延长线交于点E,∠ODA=∠OBC,AD=CB,求证:AE=CE

2025-06-26 05:46:16
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回答1:

证明:在△AOD与△COB中,

∠ODA=∠OBC
∠AOD=∠COB
AD=BC

∴△AOD≌△COB(AAS);
∴∠A=∠C,OA=OC,OD=OB,
∴OA+OB=OC+OD,即AB=CD.
∵在△ABE与△CDE中,
∠E=∠E
∠A=∠C
AB=CD

∴△ABE≌△CDE(AAS),
∴AE=CE.