题目内容
阅读材料如图①,△ABC与△DEF都是等腰直角三角形,∠ACB=∠EDF=90°,且点D在AB边上,AB、EF的中点均为O,连结BF、CD、CO,显然点C、F、O在同一条直线上,可以证明△BOF≌△COD,则BF=CD.
解决问题
(1)将图①中的Rt△DEF绕点O旋转得到图②,猜想此时线段BF与CD的数量关系,并证明你的结论;
(2)如图③,若△ABC与△DEF都是等边三角形,AB、EF的中点均为O,上述(1)中的结论仍然成立吗?如果成立,请说明理由;如不成立,请求出BF与CD之间的数量关系;
(3)如图④,若△ABC与△DEF都是等腰三角形,AB、EF的中点均为0,且顶角∠ACB=∠EDF=α,请直接写出
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_ST/images1.png)
【答案】分析:(1)如答图②所示,连接OC、OD,证明△BOF≌△COD;
(2)如答图③所示,连接OC、OD,证明△BOF∽△COD,相似比为
;
(3)如答图④所示,连接OC、OD,证明△BOF∽△COD,相似比为tan
.
解答:解:(1)猜想:BF=CD.理由如下:
如答图②所示,连接OC、OD.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/images2.png)
∵△ABC为等腰直角三角形,点O为斜边AB的中点,
∴OB=OC,∠BOC=90°.
∵△DEF为等腰直角三角形,点O为斜边EF的中点,
∴OF=OD,∠DOF=90°.
∵∠BOF=∠BOC+∠COF=90°+∠COF,∠COD=∠DOF+∠COF=90°+∠COF,
∴∠BOF=∠COD.
∵在△BOF与△COD中,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/2.png)
∴△BOF≌△COD(SAS),
∴BF=CD.
(2)答:(1)中的结论不成立.
如答图③所示,连接OC、OD.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/images4.png)
∵△ABC为等边三角形,点O为边AB的中点,
∴
=tan30°=
,∠BOC=90°.
∵△DEF为等边三角形,点O为边EF的中点,
∴
=tan30°=
,∠DOF=90°.
∴
=
=
.
∵∠BOF=∠BOC+∠COF=90°+∠COF,∠COD=∠DOF+∠COF=90°+∠COF,
∴∠BOF=∠COD.
在△BOF与△COD中,
∵
=
=
,∠BOF=∠COD,
∴△BOF∽△COD,
∴
=
.
(3)如答图④所示,连接OC、OD.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/images17.png)
∵△ABC为等腰三角形,点O为底边AB的中点,
∴
=tan
,∠BOC=90°.
∵△DEF为等腰三角形,点O为底边EF的中点,
∴
=tan
,∠DOF=90°.
∴
=
=tan
.
∵∠BOF=∠BOC+∠COF=90°+∠COF,∠COD=∠DOF+∠COF=90°+∠COF,
∴∠BOF=∠COD.
在△BOF与△COD中,
∵
=
=tan
,∠BOF=∠COD,
∴△BOF∽△COD,
∴
=tan
.
点评:本题是几何综合题,考查了旋转变换中相似三角形、全等三角形的判定与性质.解题关键是:第一,善于发现几何变换中不变的逻辑关系,即△BOF≌△COD或△BOF∽△COD;第二,熟练运用等腰直角三角形、等边三角形、等腰三角形的相关性质.本题(1)(2)(3)问的解题思路一脉相承,由特殊到一般,有利于同学们进行学习与探究.
(2)如答图③所示,连接OC、OD,证明△BOF∽△COD,相似比为
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/0.png)
(3)如答图④所示,连接OC、OD,证明△BOF∽△COD,相似比为tan
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/1.png)
解答:解:(1)猜想:BF=CD.理由如下:
如答图②所示,连接OC、OD.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/images2.png)
∵△ABC为等腰直角三角形,点O为斜边AB的中点,
∴OB=OC,∠BOC=90°.
∵△DEF为等腰直角三角形,点O为斜边EF的中点,
∴OF=OD,∠DOF=90°.
∵∠BOF=∠BOC+∠COF=90°+∠COF,∠COD=∠DOF+∠COF=90°+∠COF,
∴∠BOF=∠COD.
∵在△BOF与△COD中,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/2.png)
∴△BOF≌△COD(SAS),
∴BF=CD.
(2)答:(1)中的结论不成立.
如答图③所示,连接OC、OD.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/images4.png)
∵△ABC为等边三角形,点O为边AB的中点,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/4.png)
∵△DEF为等边三角形,点O为边EF的中点,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/6.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/9.png)
∵∠BOF=∠BOC+∠COF=90°+∠COF,∠COD=∠DOF+∠COF=90°+∠COF,
∴∠BOF=∠COD.
在△BOF与△COD中,
∵
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/10.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/12.png)
∴△BOF∽△COD,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/13.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/14.png)
(3)如答图④所示,连接OC、OD.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/images17.png)
∵△ABC为等腰三角形,点O为底边AB的中点,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/15.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/16.png)
∵△DEF为等腰三角形,点O为底边EF的中点,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/17.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/18.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/19.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/20.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/21.png)
∵∠BOF=∠BOC+∠COF=90°+∠COF,∠COD=∠DOF+∠COF=90°+∠COF,
∴∠BOF=∠COD.
在△BOF与△COD中,
∵
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/22.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/23.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/24.png)
∴△BOF∽△COD,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/25.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131101193716221723070/SYS201311011937162217230026_DA/26.png)
点评:本题是几何综合题,考查了旋转变换中相似三角形、全等三角形的判定与性质.解题关键是:第一,善于发现几何变换中不变的逻辑关系,即△BOF≌△COD或△BOF∽△COD;第二,熟练运用等腰直角三角形、等边三角形、等腰三角形的相关性质.本题(1)(2)(3)问的解题思路一脉相承,由特殊到一般,有利于同学们进行学习与探究.
![](http://thumb2018.1010pic.com/images/loading.gif)
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