已知如图,点D是角ABC的角平分线与角ACB的外 - 知识问答

已知如图,点D是角ABC的角平分线与角ACB的外角平分线的交点,DE平分BC,DE交AB于点E

1个回答

∵BD平分∠ABC
∴∠ABD = ∠DBC
∵DE //BC
∴∠EDB = ∠DBC
∴∠ABD = ∠EDB
∴EB = ED
∵CD是∠ACG的平分线
∴∠ACD = ∠DCG
∵ED//BC
∴∠FDC = ∠DCG
∴∠ACD = ∠FDC
∴FC = FD
∵EF = ED-FD
∴EF = EB - FC

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