(2014?海淀区一模)如图,在△ABC中,AB=AC,以AB为直径的⊙O与边BC、AC分别交于D、E两点,DF⊥AC于F.

2025-03-14 21:36:25
推荐回答(2个)
回答1:

(1)连接OD,AD,
∵AB是⊙的直径,
∴∠ADB=90°,
又∵AB=AC,
∴BD=CD
又∵OB=OA,
∴OD∥AC
∵DF⊥AC,
∴OD⊥DF
又∵OD为⊙的半径,
∴DF为⊙O的切线.

(2)连接BE交OD于M,过O作ON⊥AE于N,
则AE=2NE,
∵cosC=
3
5
,CF=9,
∴DC=15,
∴DF=
152?92
=12,
∵AB是直径,
∴∠AEB=∠CEB=90°,
∵DF⊥AC,OD⊥DF,
∴∠DFE=∠FEM=∠MDF=90°,
∴四边形DMEF是矩形,
∴EM=DF=12,∠DME=90°,DM=EF,
即OD⊥BE,
同理四边形OMEN是矩形,
∴OM=EN,
∵OD为半径,
∴BE=2EM=24,
∵∠BEA=∠DFC=90°,∠C=∠C,
∴△CFD∽△CEB,
DF
BE
=
CF
CE

12
24
=
9
9+EF

∴EF=9=DM,
设⊙O的半径为R,
则在Rt△EMO中,由勾股定理得:R2=122+(R-9)2
解得:R=
225
18

则EN=OM=
225
18
-9=
63
18
=
7
2

∴AE=2EN=7.

回答2:

(1)证明:连接OD
因为AB=AC
所以三角形ABC是等腰三角形
因为AB是圆O的直径
所以角ADB=90度
AD是等腰三角形ABC的垂线,角平分线
所以角BAD=1/2角BAC
因为角BOC=弧BD
角BAD=1/2弧BD
俗语角BOC=角BAC
所以OD平行AC
所以角ODF=角CFD
因为DF垂直AC
所以角CFD=90度
所以角ODF=90度
所以半径OD垂直DF
所以DF为圆O的切线
(2)解:连接DE
因为AB=AC
所以角B=角C
因为角CED=角B
所以角C=角CED
所以CD=ED
所以三角形CDE是等腰三角形
因为DF垂直AC
所以DF是等腰三角形CDF的垂线,中线
所以CF=EF=1/2CE
角AFD=角CFD=90度(已证)
所以三角形CFD和三角形AFD是直角三角形
所以CF^2+DF^2=CD^2
cosC=DF/CD=3/5
tanC=DF/CF
tan角ADF=AF/DF
所以tanC=4/3
因为CF=9
所以DF=12
EF=9
因为DF是圆O的切线
所以角ADF=角B
因为角B=角C(已证)
所以角ADF=角C
所以tanC=tan角ADF=AF/DF=4/3
所以AF=14
因为AF=AE+EF
所以AE=16-12=4
所以AE的长是4

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