Question

如图，已知D是∠ABC的平分线与∠ACB的外角平分线的交点，DE∥BC，交AB于E，交AC于F。求证：EF=BE-CF。

没图啊！~~要是哪位仁兄做过这道题的证明一下吧~~证明得好，分追加啊~~

Answer 1

∵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

Answer 2

因为DE∥BC
所以∠DBC=∠EDB ,∠DCG=∠EDC
又因为两个角平分线
∠EBD=∠DBC，∠FCD=∠DCG
所以等价代换∠EBD=∠EDB所以EB=ED
∠FCD=∠EDC所以CF=FD
EF=ED-FD
所以EF=EB-CF

Answer 3

BE=DE
CF=DF
EF=DE-DF=BE-CF

Answer 4

∵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

Answer 5

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