条 件:如下左图,A、B是直线同旁的两个定点.
问 题:在直线l上确定一点P,使PA+PB的值最小.
方 法:作点A关于直线l的对称点A',连结A'B交l于点P,则PA+PB=A'B的值最小(不必证明).
模型应用:
(1)如图1,正方形ABCD的边长为2,E为AB的中点,P是AC上一动点.连结BD,由正方形对称性可知,B与D关于直线AC对称.连结PE、PB,则PB+PE的最小值是( );
(2)如图2,
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20101116/20101116144137406888.gif)
的半径为2,点A、B、C在
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20101116/20101116144144031888.gif)
上,
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20101116/20101116144150125955.gif)
,
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20101116/20101116144156421986.gif)
,P是OB上一动点,求PA+PC的最小值;
(3)如图3,∠AOB=30°,P是
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20101116/20101116144215078927.gif)
内一点,PO=8,Q,R分别是OA、OB上的动点,求
![](http://thumb.1010pic.com/pic1/upload/papers/c02/20101116/20101116144225140951.gif)
周长的最小值.