【答案】
分析:(1)①先根據(jù)等腰三角形等角對(duì)等邊的性質(zhì)及三角形內(nèi)角和定理得出∠DAE=∠BAC,則∠BAD=∠CAE,再根據(jù)SAS證明△ABD≌△ACE,從而得出BD=CE;
②先由全等三角形的對(duì)應(yīng)角相等得出∠BDA=∠CEA,再根據(jù)三角形的外角性質(zhì)即可得出∠BMC=∠DAE=180°-2α;
(2)先根據(jù)等腰三角形等角對(duì)等邊的性質(zhì)及三角形內(nèi)角和定理得出∠DAE=∠BAC=90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/0.png)
α,則∠BAD=∠CAE,再由AB=kAC,AD=kAE,得出AB:AC=AD:AE=k,則根據(jù)兩邊對(duì)應(yīng)成比例,且?jiàn)A角相等的兩三角形相似證出△ABD∽△ACE,得出BD=kCE,∠BDA=∠CEA,然后根據(jù)三角形的外角性質(zhì)即可得出∠BMC=∠DAE=90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/1.png)
α;
(3)先在備用圖中利用SSS作出旋轉(zhuǎn)后的圖形,再根據(jù)等腰三角形等角對(duì)等邊的性質(zhì)及三角形內(nèi)角和定理得出∠DAE=∠BAC=90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/2.png)
α,由AB=kAC,AD=kAE,得出AB:AC=AD:AE=k,從而證出△ABD∽△ACE,得出∠BDA=∠CEA,然后根據(jù)三角形的外角性質(zhì)即可得出∠BMC=90°+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/3.png)
α.
解答:解:(1)如圖1.
①BD=CE,理由如下:
∵AD=AE,∠ADE=α,
∴∠AED=∠ADE=α,
∴∠DAE=180°-2∠ADE=180°-2α,
同理可得:∠BAC=180°-2α,
∴∠DAE=∠BAC,
∴∠DAE+∠BAE=∠BAC+∠BAE,
即:∠BAD=∠CAE.
在△ABD與△ACE中,
∵
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/4.png)
,
∴△ABD≌△ACE(SAS),
∴BD=CE;
②∵△ABD≌△ACE,
∴∠BDA=∠CEA,
∵∠BMC=∠MCD+∠MDC,
∴∠BMC=∠MCD+∠CEA=∠DAE=180°-2α;
(2)如圖2.
∵AD=ED,∠ADE=α,
∴∠DAE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/5.png)
=90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/6.png)
α,
同理可得:∠BAC=90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/7.png)
α,
∴∠DAE=∠BAC,
∴∠DAE+∠BAE=∠BAC+∠BAE,
即:∠BAD=∠CAE.
∵AB=kAC,AD=kAE,
∴AB:AC=AD:AE=k.
在△ABD與△ACE中,
∵AB:AC=AD:AE=k,∠BDA=∠CEA,
∴△ABD∽△ACE,
∴BD:CE=AB:AC=AD:AE=k,∠BDA=∠CEA,
∴BD=kCE;
∵∠BMC=∠MCD+∠MDC,
∴∠BMC=∠MCD+∠CEA=∠DAE=90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/8.png)
α.
故答案為:BD=kCE,90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/9.png)
α;
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/images10.png)
(3)如右圖.
∵AD=ED,∠ADE=α,
∴∠DAE=∠AED=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/10.png)
=90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/11.png)
α,
同理可得:∠BAC=90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/12.png)
α,
∴∠DAE=∠BAC,即∠BAD=∠CAE.
∵AB=kAC,AD=kAE,
∴AB:AC=AD:AE=k.
在△ABD與△ACE中,
∵AB:AC=AD:AE=k,∠BAD=∠CAE,
∴△ABD∽△ACE,
∴∠BDA=∠CEA,
∵∠BMC=∠MCD+∠MDC,∠MCD=∠CED+∠ADE=∠CED+α,
∴∠BMC=∠CED+α+∠CEA=∠AED+α=90°-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/13.png)
α+α=90°+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/14.png)
α.
故答案為:90°+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101190208887744528/SYS201311011902088877445024_DA/15.png)
α.
點(diǎn)評(píng):本題考查了全等三角形的判定與性質(zhì),三角形的外角的性質(zhì),相似三角形的判定與性質(zhì),作圖-旋轉(zhuǎn)變換,綜合性較強(qiáng),有一定難度.由于全等是相似的特殊情況,所以做第二問(wèn)可以借助第一問(wèn)的思路及方法,做第三問(wèn)又可以遵照第二問(wèn)的做法,本題三問(wèn)由淺入深,層層遞進(jìn),做好第一問(wèn)是關(guān)鍵.