![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
如图,△abc中,ab=ac,∠bac=90°,cd平分∠acb,be⊥cd,垂足e在cd的延长线
1个回答
2013-11-01 · 知道合伙人软件行家
关注
![](https://wyw-base.cdn.bcebos.com/pc-content/follow.gif)
展开全部
解:延长CD交BM的延长线于F.
∠FBD=∠CBD,
BD=BD
∠BDF=∠BDC=90°
则⊿BDF≌⊿BDC
BF=BC;DF=DC.
DM与CA都垂直于BF
则:DM∥CA
FM/MA=DF/DC=1
FM=MA.
∴BM/(AB+BC)=BM/[(BM-AM)+(BM+FM)]=BM/(2BM)=1/2;
AM/(BC-AB)=AM/(BF-AB)=AM/(2AM)=1/2.
∠FBD=∠CBD,
BD=BD
∠BDF=∠BDC=90°
则⊿BDF≌⊿BDC
BF=BC;DF=DC.
DM与CA都垂直于BF
则:DM∥CA
FM/MA=DF/DC=1
FM=MA.
∴BM/(AB+BC)=BM/[(BM-AM)+(BM+FM)]=BM/(2BM)=1/2;
AM/(BC-AB)=AM/(BF-AB)=AM/(2AM)=1/2.
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询