解:
(1)∵AE⊥EF, EF∥BC,∴AD⊥BC. (1分)
在△ABD和△ACD中,∵BD=CD,∠ADB=∠ADC,AD=AD,
∴△ABD≌△ACD.(或者:又∵BD=CD,∴AE是BC的中垂线.) (2分)
∴AB=AC. (3分)
(2)连BO,∵AD是BC的中垂线,∴BO=CO.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082300455271072.gif)
(或者:证全等也可得到BO=CO.)
又AO=CO,∴AO=BO=CO. (4分)
∴点O是△ABC外接圆的圆心. (5分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408230045527265297.jpg)
(3)解法1:
∵∠ABE=∠ADB=90°,∴∠ABD+∠BAD=∠AEB+∠BAE=90
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082300455274185.gif)
°,
∴∠ABD=∠AEB. 又∵∠BAD=∠EAB, ∴△ABD∽△AEB.
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823004552757589.gif)
(或者:由三角函数得到
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823004552757589.gif)
) (6分)
在Rt△ABD中,∵AB=5,BD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823004552788225.gif)
BC=3, ∴AD=4. (7分)
∴AE=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823004552804251.gif)
. (8分)
解法2:
∵AO=BO, ∴∠ABO=∠BAO.
∵∠ABE=90°,∴∠ABO+∠OBE=∠BAO+∠AEB=90°.
∴∠OBE=∠OEB, ∴OB=OE. (6分)
在 Rt△ABD中,∵AB=5,BD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823004552788225.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082300455283572.gif)
BC=3,∴AD=4.
设 OB=x, 则 OD=4-x,由3
2+(4-x)
2=x
2,解得x=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823004552850263.gif)
. (7分)
∴AE=2OB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823004552804251.gif)
.(8分)
解法3:
设AO
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082300455288285.gif)
的延长线与⊙O交于点E
1,则AE
1是⊙O的直径, ∴∠ABE
1=90°.
在Rt△ABE和Rt△ABE
1中,∵∠BAE=∠BAE
1,∠ABE=∠ABE
1=90
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082300455289771.gif)
°,AB=AB,
∴△ABE≌△ABE
1,∴AE=AE
1. (6分) (同方法2) ∵BO=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823004552850263.gif)
. (7分)
∴AE=2OB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823004552804251.gif)
. (8分)