如图,D是三角形ABC的内心,且CD+AD=BC,且角BAC=80度,求角ABC和角ADB的度数?【“内心”即角平分线的交点】
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 18:19:33
![如图,D是三角形ABC的内心,且CD+AD=BC,且角BAC=80度,求角ABC和角ADB的度数?【“内心”即角平分线的交点】](/uploads/image/z/1677899-11-9.jpg?t=%E5%A6%82%E5%9B%BE%EF%BC%8CD%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%86%85%E5%BF%83%EF%BC%8C%E4%B8%94CD%2BAD%3DBC%EF%BC%8C%E4%B8%94%E8%A7%92BAC%3D80%E5%BA%A6%EF%BC%8C%E6%B1%82%E8%A7%92ABC%E5%92%8C%E8%A7%92ADB%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%9F%E3%80%90%E2%80%9C%E5%86%85%E5%BF%83%E2%80%9D%E5%8D%B3%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E7%9A%84%E4%BA%A4%E7%82%B9%E3%80%91)
如图,D是三角形ABC的内心,且CD+AD=BC,且角BAC=80度,求角ABC和角ADB的度数?【“内心”即角平分线的交点】
如图,D是三角形ABC的内心,且CD+AD=BC,且角BAC=80度,求角ABC和角ADB的度数?【“内心”即角平分线的交点】
如图,D是三角形ABC的内心,且CD+AD=BC,且角BAC=80度,求角ABC和角ADB的度数?【“内心”即角平分线的交点】
做△ABC的外接圆,延长CD交圆于E,连接AE、BE,(∠B+∠C)/2=50°,∠BDE=∠DBE=50°,△BED为等腰△,BE=DE,∠ADE=(∠A+∠C)/2=40°+∠C/2=∠DAE,△AED为等腰△,AE=DE,在CB上取CF=CD,连接DF,∠CFD=(180°-∠C/2)/2,∠BFD=180°-(180°-∠C/2)/2=90°+∠C/4,∠BDC=130°,∠BDF=130°-(180°-∠C/2)/2=40°+∠C/4,AE=DE=AD/2cos(40°+∠C/2),BD=2DEcos50°=ADcos50°/cos(40°+∠C/2),CD+AD=BC,则AD=BF,在△BDF中由正弦定理得:BF/sin(40°+∠C/4)=BD/sin(90°+∠C/4),1/sin(40°+∠C/4)=cos50°/[cos(40°+∠C/2)sin(90°+∠C/4)],sin(40°+∠C/4)sin40°=cos(40°+∠C/2)cos(∠C/4),[-cos(80°+∠C/4)+cos(∠C/4)]/2=cos(40°+∠C/2)cos(∠C/4),[-cos(80°+∠C/4)]/2=cos(∠C/4)[-1/2+cos(40°+∠C/2)]=cos(∠C/4)[cos120°+cos(40°+∠C/2)]=2cos(80°+∠C/4)cos(40°-∠C/4)],cos(80°+∠C/4)[1+2cos(40°-∠C/4)]=0,1+2cos(40°-∠C/4)≠0,取cos(80°+∠C/4)=0,80°+∠C/4=90°,∠C=40°,∠B=∠AED=60°,△AED为等边△,∠ADB=60°+50°=110°.