![](http://thumb.1010pic.com/pic5/upload/201308/528399217f326.png)
证明:连接OE、OF、AE、AF,延长SA交FE的延长线于K,
∵
![](http://thumb.1010pic.com/pic5/latex/24375.png)
=
![](http://thumb.1010pic.com/pic5/latex/209675.png)
,
∴AB∥EF,
∴
![](http://thumb.1010pic.com/pic5/latex/544590.png)
=
![](http://thumb.1010pic.com/pic5/latex/544591.png)
=
![](http://thumb.1010pic.com/pic5/latex/206950.png)
,
∵AC=CD,
∴KE=EF=AE,∠KAF=90°,
FA⊥SA,又
![](http://thumb.1010pic.com/pic5/latex/24375.png)
=
![](http://thumb.1010pic.com/pic5/latex/15226.png)
,
∴OE⊥FA,OE∥SA,
同理可证OF∥SB,
∴∠ASB=∠EOF=
![](http://thumb.1010pic.com/pic5/latex/8.png)
∠AOB.
分析:可连接OE、OF、AE、AF,延长SA交FE的延长线于K,由于E、F是弧AB的三等分点,即
![](http://thumb.1010pic.com/pic5/latex/24375.png)
=
![](http://thumb.1010pic.com/pic5/latex/209675.png)
,得出AB∥EF,进而得出对应线段成比例,由线段之间的关系得出∠KAF=90°,得出OE∥SA,以及OF∥SB,进而可求解结论.
点评:本题主要考查了圆心角、弦及弧之间的关系和平行线分线段成比例的性质问题,能够熟练掌握.