如图,点A,B,C,D在圆O上,AB=AC,AD与BC相交于点E,AE=ED/2,延长DB到点F,使FB=BD/2,连接AF
求证△ABE∽△ADB;求证直线AF与圆O相切.
人气:477 ℃ 时间:2019-08-20 09:29:13
解答
1、∵AB=AC
∴∠ABC=∠ACB
∵∠ACB=∠BDA(同弧上圆周角相等)
∴∠ABC=∠BDA
即∠ABE=∠BDA
∵∠BAE=∠BAD
∴△ABE∽△ADB
2、∵AE=ED/2即AE/ED=1/2
FB=BD/2即FB/BD=1/2
∴AE/ED=FB/BD
∴(AE+ED)/ED=(FB+BD)/BD
即AD/ED=BD/DF
∵∠ADF=∠EDB
∴△ADF∽△EDB
∴∠EBD=∠AFD
∴AF∥BE
∴∠FAB=∠ABE=∠BDA
即∠FAB=∠FDA
∵∠AFB=∠AFD
∴△AFB∽△ADF
∴AF/DF=BF/AF
即AF²=BF×DF
∴直线AF与圆O相切(切割线逆定理)
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