这道题不理解请老师帮助
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/30 15:00:40
![](http://img.wesiedu.com/upload/a/a3/aa3a8a041beac851bcb12b3833593f87.png)
![](http://img.wesiedu.com/upload/7/73/7733485460004dc9116d128db5b1e192.png)
这道题不理解请老师帮助
![这道题不理解请老师帮助](/uploads/image/z/10800073-1-3.jpg?t=%E8%BF%99%E9%81%93%E9%A2%98%E4%B8%8D%E7%90%86%E8%A7%A3%E8%AF%B7%E8%80%81%E5%B8%88%E5%B8%AE%E5%8A%A9)
解题思路: 根据等腰三角形的性质,线段垂直平分线的性质,可得△ODC是等腰三角形,先根据等腰直角三角形的性质和勾股定理得到AC,BC,OB,OA,OC,AD,OD,CD,BD的长度,再根据相似三角形的判定与性质分两种情况得到BM的长度,进一步得到点M的坐标.
解题过程:
解:∵OB=CB,OB边上的高CA与OC边上的高BE相交于点D,AB=
,∠CBO=45°,
∴AB=AC=
,OD=CD,∠BOC=
=67.5°,
在Rt△BAC中,BC=
=2,
∴OB=2,
∴OA=OB﹣AB=2﹣
,
在Rt△OAC中,OC=
=2
,
在Rt△OAD中,OA2+AD2=OD2,
(2﹣
)2+AD2=(
﹣AD)2,
解得:AD=2﹣
,
∴OA=AD,∠DOA=45°,
∴OD=CD=2
﹣2,
在Rt△BAD中,BD=
=2
,
①如图1,△BMC∽△CDO时,过M点作MF⊥AB于F,
![](http://img.wesiedu.com/upload/c/6b/c6b34a37c44f74fc038a7d9730e6d483.png)
=
,即
=
,
解得BM=
,
∵MF⊥AB,CA是OB边上的高,
∴MF∥DA,
∴△BMF∽△BDA,
∴
=
=
,即
=
=
,
解得BF=1,MF=
﹣1,
∴OF=OB﹣BF=1,
∴点M的坐标是(1,
﹣1);
②如图2,△BCM∽△CDO时,过M点作MF⊥AB于F,
![](http://img.wesiedu.com/upload/d/60/d607ec58480e82ff73e7f76c169827b0.png)
=
,即
=
,
解得BM=2
,
∵MF⊥AB,CA是OB边上的高,
∴MF∥DA,
∴△BMF∽△BDA,
∴
=
=
,即
=
=
,
解得BF=2+
,MF=
,
∴OF=BF﹣OB=
,
∴点M的坐标是(﹣
,
).
综上所述,点M的坐标是(1,
﹣1)或(﹣
,
).
故答案为:(1,
﹣1)或(﹣
,
).
解题过程:
解:∵OB=CB,OB边上的高CA与OC边上的高BE相交于点D,AB=
![](http://img.wesiedu.com/upload/3/5e/35e3df3c5c95c5fa266f5325a8c2c441.png)
∴AB=AC=
![](http://img.wesiedu.com/upload/3/5e/35e3df3c5c95c5fa266f5325a8c2c441.png)
![](http://img.wesiedu.com/upload/6/ef/6efe7728a20466c0cfe82a6d33887942.png)
在Rt△BAC中,BC=
![](http://img.wesiedu.com/upload/f/89/f89ad3b5e6c4e31c2731f3f13e7a4f51.png)
∴OB=2,
∴OA=OB﹣AB=2﹣
![](http://img.wesiedu.com/upload/3/5e/35e3df3c5c95c5fa266f5325a8c2c441.png)
在Rt△OAC中,OC=
![](http://img.wesiedu.com/upload/2/ca/2ca20bd9a25692ad6b41e7360a2162ab.png)
![](http://img.wesiedu.com/upload/d/5f/d5fecd89e2d535a18a03384848c3ebbc.png)
在Rt△OAD中,OA2+AD2=OD2,
(2﹣
![](http://img.wesiedu.com/upload/3/5e/35e3df3c5c95c5fa266f5325a8c2c441.png)
![](http://img.wesiedu.com/upload/3/5e/35e3df3c5c95c5fa266f5325a8c2c441.png)
解得:AD=2﹣
![](http://img.wesiedu.com/upload/3/5e/35e3df3c5c95c5fa266f5325a8c2c441.png)
∴OA=AD,∠DOA=45°,
∴OD=CD=2
![](http://img.wesiedu.com/upload/3/5e/35e3df3c5c95c5fa266f5325a8c2c441.png)
在Rt△BAD中,BD=
![](http://img.wesiedu.com/upload/d/b6/db6451569428b6df44146d1d54656a63.png)
![](http://img.wesiedu.com/upload/4/9d/49d7c954e15b010f0429e965e49da4b9.png)
①如图1,△BMC∽△CDO时,过M点作MF⊥AB于F,
![](http://img.wesiedu.com/upload/c/6b/c6b34a37c44f74fc038a7d9730e6d483.png)
![](http://img.wesiedu.com/upload/f/6e/f6e78953a23c6d0f33187f9086e32092.png)
![](http://img.wesiedu.com/upload/f/29/f2996a3d0a5a6c6d225cb00acb951e87.png)
![](http://img.wesiedu.com/upload/c/99/c99217ff786fd7cfcdfd2bb9d1c9b17c.png)
![](http://img.wesiedu.com/upload/8/db/8db60ac16f1e7f154b9ab7cc9cd87884.png)
解得BM=
![](http://img.wesiedu.com/upload/9/59/959c6abb8a1ac480ba6536b2ba9a849c.png)
∵MF⊥AB,CA是OB边上的高,
∴MF∥DA,
∴△BMF∽△BDA,
∴
![](http://img.wesiedu.com/upload/a/7e/a7e58601e2707a0a792dd6ab54fcc7ab.png)
![](http://img.wesiedu.com/upload/d/d7/dd7dabb072a0d9eeb6f6b56d61cf8f13.png)
![](http://img.wesiedu.com/upload/6/d5/6d5391014ae49d05eb47d4aa74649d67.png)
![](http://img.wesiedu.com/upload/8/77/8775d4404157509a8cc011d8225430be.png)
![](http://img.wesiedu.com/upload/6/21/621fe20493b3002e33b01d581c43cad4.png)
![](http://img.wesiedu.com/upload/5/68/568be05a50a9c600632b3a7f59c8b1e2.png)
解得BF=1,MF=
![](http://img.wesiedu.com/upload/5/17/517785213c57632c2ca825dc3d0eea94.png)
∴OF=OB﹣BF=1,
∴点M的坐标是(1,
![](http://img.wesiedu.com/upload/5/17/517785213c57632c2ca825dc3d0eea94.png)
②如图2,△BCM∽△CDO时,过M点作MF⊥AB于F,
![](http://img.wesiedu.com/upload/d/60/d607ec58480e82ff73e7f76c169827b0.png)
![](http://img.wesiedu.com/upload/7/69/769c521d58ded257b269ecbe0fd1cfbf.png)
![](http://img.wesiedu.com/upload/0/da/0dabab0cd69c247ff7d053ca60e06080.png)
![](http://img.wesiedu.com/upload/9/98/9980a93c67eb19053fced407e94e8410.png)
![](http://img.wesiedu.com/upload/2/cb/2cbda64f62d9685cfb540df091509f1e.png)
解得BM=2
![](http://img.wesiedu.com/upload/3/80/3809b567a0084379ad021955847fb1c5.png)
∵MF⊥AB,CA是OB边上的高,
∴MF∥DA,
∴△BMF∽△BDA,
∴
![](http://img.wesiedu.com/upload/7/c2/7c294b8b45b6815c1fa52520d8a5f869.png)
![](http://img.wesiedu.com/upload/e/ce/ece8565913eda265ad9db933e5d46270.png)
![](http://img.wesiedu.com/upload/2/2c/22c3fef3f68014caaf77e0e0cac66be0.png)
![](http://img.wesiedu.com/upload/8/42/84288fbb2ed4f8187b97648ac4c7a58a.png)
![](http://img.wesiedu.com/upload/9/ab/9ab6cf562d6dbe1adf0a81a8e710af90.png)
![](http://img.wesiedu.com/upload/0/bb/0bbcbc9b6664631ae4b73132a1a327fa.png)
解得BF=2+
![](http://img.wesiedu.com/upload/8/8a/88abc699478b89e9a31441231d13abf1.png)
![](http://img.wesiedu.com/upload/8/8a/88abc699478b89e9a31441231d13abf1.png)
∴OF=BF﹣OB=
![](http://img.wesiedu.com/upload/8/8a/88abc699478b89e9a31441231d13abf1.png)
∴点M的坐标是(﹣
![](http://img.wesiedu.com/upload/8/8a/88abc699478b89e9a31441231d13abf1.png)
![](http://img.wesiedu.com/upload/8/8a/88abc699478b89e9a31441231d13abf1.png)
综上所述,点M的坐标是(1,
![](http://img.wesiedu.com/upload/8/8a/88abc699478b89e9a31441231d13abf1.png)
![](http://img.wesiedu.com/upload/e/dd/edd2a9f0b7b600270231f060391700f1.png)
![](http://img.wesiedu.com/upload/e/dd/edd2a9f0b7b600270231f060391700f1.png)
故答案为:(1,
![](http://img.wesiedu.com/upload/e/dd/edd2a9f0b7b600270231f060391700f1.png)
![](http://img.wesiedu.com/upload/e/dd/edd2a9f0b7b600270231f060391700f1.png)
![](http://img.wesiedu.com/upload/e/dd/edd2a9f0b7b600270231f060391700f1.png)