如图在△abc中,∠a=∠abc,直线ef分别交△abc的边ab,ac和cb的延长线于点d,e,f 求证：∠f+∠fe

学习时间:2026-08-23 10:27:22 阅读：9904

如图在△abc中,∠a=∠abc,直线ef分别交△abc的边ab,ac和cb的延长线于点d,e,f 求证：∠f+∠fec∠a

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2026-08-23 10:27:22

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∵∠C=180°-(∠F+∠FEC)∠C=180°-(∠A+∠ABC)∴180°-(∠F+∠FEC)=180°-(∠A+∠ABC)∴∠F+∠FEC=∠A+∠ABC∵∠A=∠ABC∴∠F+∠FEC =2∠A 再问：如何证出∠a=∠abc 再答： ∠a=∠abc是已知

如图 在△abc中,∠a=∠abc,直线ef分别交△abc的边ab,ac和cb的延长线于点d,e,f 求证：∠f+∠fe