A
分析:连结OC,在Rt△DCE中利用cosD=
![](http://thumb.1010pic.com/pic5/latex/284724.png)
=
![](http://thumb.1010pic.com/pic5/latex/602.png)
,可设DE=4x,则DC=5x,于是CE=3x=8,解得x=
![](http://thumb.1010pic.com/pic5/latex/137.png)
得到DE=
![](http://thumb.1010pic.com/pic5/latex/2251.png)
,DC=
![](http://thumb.1010pic.com/pic5/latex/11801.png)
,根据圆周角定理AB为直径得到∠ACB=90°,利用∠A=∠BCD可得到∠OCD=90°,在Rt△OCD中,根据cosD=
![](http://thumb.1010pic.com/pic5/latex/48383.png)
=
![](http://thumb.1010pic.com/pic5/latex/602.png)
=
![](http://thumb.1010pic.com/pic5/latex/432741.png)
,解得OD=
![](http://thumb.1010pic.com/pic5/latex/14906.png)
,则OE=OD-DE=6,接着根据勾股定理计算出OC,然后再次利用勾股定理计算AC.
解答:连结OC,如图,
![](http://thumb.1010pic.com/pic5/upload/201310/5284fd4c53420.png)
∵CE⊥AB,
∴∠AEC=∠CED=90°,
∴cosD=
![](http://thumb.1010pic.com/pic5/latex/284724.png)
=
![](http://thumb.1010pic.com/pic5/latex/602.png)
,
设DE=4x,则DC=5x,
∴CE=3x=8,解得x=
![](http://thumb.1010pic.com/pic5/latex/137.png)
,
∴DE=
![](http://thumb.1010pic.com/pic5/latex/2251.png)
,DC=
![](http://thumb.1010pic.com/pic5/latex/11801.png)
,
∵AB为直径,
∴∠ACB=90°,
∵∠A=∠BCD,
而∠A=∠ACO,
∴∠ACO=∠BCD,
∴∠OCD=90°,
在Rt△OCD中,cosD=
![](http://thumb.1010pic.com/pic5/latex/48383.png)
=
![](http://thumb.1010pic.com/pic5/latex/602.png)
=
![](http://thumb.1010pic.com/pic5/latex/432741.png)
,解得OD=
![](http://thumb.1010pic.com/pic5/latex/14906.png)
,
∴OE=OD-DE=
![](http://thumb.1010pic.com/pic5/latex/14906.png)
-
![](http://thumb.1010pic.com/pic5/latex/2251.png)
=6,
在Rt△OCE中,OC=
![](http://thumb.1010pic.com/pic5/latex/275095.png)
=10,
∴OA=10,
∴AE=10+6=16,
在Rt△ACE中,AC=
![](http://thumb.1010pic.com/pic5/latex/214220.png)
=
![](http://thumb.1010pic.com/pic5/latex/260106.png)
=8
![](http://thumb.1010pic.com/pic5/latex/559.png)
.
故选A.
点评:本题考查了相似三角形的判定与性质:两组角对应相等的两个三角形相似;相似三角形对应边的比相等.也考查了圆周角定理和解直角三角形.