archise
花朵
共回答了14个问题采纳率:100% 举报
(1)联结AD
∵AB是⊙O的直径,∴∠ADB=∠AEB =90° --- 1分
∵AB=AC,∴CD=BD
∵OA=OB,∴OD//AC
∴OD⊥BE--------------------------------------- 2分
(2)方法一:∵∠CEB=∠AEB=90°,CD="BD,AB=5," DE=
∴AC="AB=5, " BC=2DE=2
![](https://img.yulucn.com/upload/e/ff/effa3da01f032c5dc231ce4573b34488_thumb.jpg)
, --------------------- 3分
在△ABE、△BCE中,∠CEB=∠AEB=90°,则有
设AE="x," 则
![](https://img.yulucn.com/upload/3/54/35439988712cea8414917369a704ff19_thumb.jpg)
--------------------- 4分
解得:x="3"
∴AE="3" -------------------------- 5分
方法二:∵OD⊥BE,∴BD=DE,BF="EF" ------------------------3分
设AE=x,∴OF=
![](https://img.yulucn.com/upload/c/24/c243a143f1c33f1051ade6ac5e317391_thumb.jpg)
,在△OBF、△BDF中,∠OFB=∠BFD=90°
∴
∵DE=
![](https://img.yulucn.com/upload/e/ff/effa3da01f032c5dc231ce4573b34488_thumb.jpg)
,AB=5,∴
![](https://img.yulucn.com/upload/6/62/662ba8013488cfbb474bdc658c684031_thumb.jpg)
------4分
解得:x=3,∴AE="3" ------------5分
方法三:∵BE⊥AC AD⊥BC,
∴S△ABC=
![](https://img.yulucn.com/upload/1/db/1dbb31cd8abb62b5162d2a2ed6f0a6e2_thumb.jpg)
BC·AD=
![](https://img.yulucn.com/upload/1/db/1dbb31cd8abb62b5162d2a2ed6f0a6e2_thumb.jpg)
AC·BE, ----------------------------3分
∴BC·AD=AC·BE
∵BC=2DE=2
![](https://img.yulucn.com/upload/e/ff/effa3da01f032c5dc231ce4573b34488_thumb.jpg)
,AC=AB=5
∴BE="4" , ----------------------------------4分
∴AE="3" ------------------------------5分
略
1年前
8