题目

如图，A与A′关于直线MN对称，P是BA′与MN的交点.若P1为直线MN上任意一点（不与P重合），连结AP1、BP1，试说明 AP1+BP1>AP+BP. 答案：【答案】见解析【解析】试题分析：由三角形三条边的关系可得A′P1+BP1>A′B，再由轴对称的性质可得AP1=A′P1，然后通过等量代换可证明结论.解：如图，连结AP1，则在△A′P1B中，有A′P1+BP1>A′B∴A′P1+BP1>A′P+PB∵A与A′关于直线MN对称，∴AP1与A′P1关于直线MN对称∴AP1=A′P1同理可得：AP=A′P∴AP1+BP1He hid himself behind the door to avoid by his brother.A. to see B. seeing C. to be seen D. being seen