![](http://hiphotos.baidu.com/zhidao/pic/item/c75c10385343fbf213022c92b37eca8065388f51.jpg)
∵AB=CD,
∴
![]() |
AB |
![]() |
CD |
∴
![]() |
AC |
![]() |
BD |
∴∠B=∠C,
∴CE=BE,
∴DE=AE=5;
如圖乙,當(dāng)點(diǎn)C在AB的右側(cè)時,同理:DE=BE=AB-AE=3,
![](http://hiphotos.baidu.com/zhidao/pic/item/b3b7d0a20cf431ad7bfcc9884836acaf2edd9851.jpg)
(2)如圖丙,若點(diǎn)A在CD的下方,連結(jié)OC,
∵AB是⊙O的直徑,AB⊥CD,
∴CE=
1 |
2 |
設(shè)OC=x,則OE=x-2,
∵AB⊥CD,
∴OE2+CE2=OC2,即(x-2)2+42=x2,
解得:x=5.
如圖丁,若點(diǎn)A在CD的上方,則AB<2AE=4,與CD=8產(chǎn)生矛盾(或與上類似地計算得OE為負(fù)數(shù)).
答:⊙O的半徑為5.