题目内容
在△ABC中,∠BAC的平分线AD交△ABC的外接圆⊙O于点E,交BC于点D,过点E作⊙O的切线交AB的延长线于点F,若AD=3
,DE=
.
求证:
(1)EF∥BC;
(2)AF=2EF.
3 |
3 |
求证:
(1)EF∥BC;
(2)AF=2EF.
证明:(1)连接OE.
∵EF切⊙O于点E,则OE⊥EF.
∵AE平分∠BAC,∴
=
.
∴OE⊥BC.
∴EF∥BC.
(2)∵EF∥BC,AD=3
,DE=
.
∴AD:DE=AB:BF=3:1.
∴BF=
AF.
∵FE是切线,FA是割线,
∴EF2=FB•FA=
FA2,
∴EF=
FA,即AF=2EF.
∵EF切⊙O于点E,则OE⊥EF.
∵AE平分∠BAC,∴
BE |
EC |
∴OE⊥BC.
∴EF∥BC.
(2)∵EF∥BC,AD=3
3 |
3 |
∴AD:DE=AB:BF=3:1.
∴BF=
1 |
4 |
∵FE是切线,FA是割线,
∴EF2=FB•FA=
1 |
4 |
∴EF=
1 |
2 |
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