等腰Rt△PAB中,∠PAB=90°,点C是AB上一点(与A、B不重合),连接PC,将线段PC绕点C顺时针旋转90°,得到线段DC.连接PD,BD.探究∠PBD的度数,以及线段AB与BD、BC的数量关系.
(1)尝试探究:如图(1),点C在线段AB上,
∵△PCD为等腰直角三角形,且∠PCD=90°,∴∠CPD=45°=∠APB,
∴∠CPD﹣∠BPC=∠APB﹣∠BPC,即∠BPD=∠APC,
又∵
,∴△PAC∽△PBD,相似比为
,∴
.
∴∠PBD= ;AB=BC+AC= .
(2)类比探索:如图(2),点C在直线AB上,且在点B右侧,还能得出与(1)中同样的结论么?请写出你得到的结论并证明
(3)拓展迁移:如图(3),点C在直线AB上,且在点A左侧,请补充完成图形,并直接写出你得到的结论(不需要证明)
(1)尝试探究:如图(1),点C在线段AB上,
∵△PCD为等腰直角三角形,且∠PCD=90°,∴∠CPD=45°=∠APB,
∴∠CPD﹣∠BPC=∠APB﹣∠BPC,即∠BPD=∠APC,
又∵
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2597d55de3c959f2e4ad3ee10f84b581.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a413c742d76e4ff850d424a7da2242db.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/606d6bc087483f2e3c2b382f0b733ecd.png)
∴∠PBD= ;AB=BC+AC= .
(2)类比探索:如图(2),点C在直线AB上,且在点B右侧,还能得出与(1)中同样的结论么?请写出你得到的结论并证明
(3)拓展迁移:如图(3),点C在直线AB上,且在点A左侧,请补充完成图形,并直接写出你得到的结论(不需要证明)
![](https://img.xkw.com/dksih/QBM/2018/12/3/2088885633482752/2095402177527808/STEM/1f70792cfc024f6c99a8eb760f7028f5.png?resizew=397)
更新时间:2018-12-12 21:29:33
|
【知识点】 相似三角形的判定与性质综合
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较难
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