如图,△ABC中,∠C=2∠b。AD是角平分线,求证:AB=AC+DC
3个回答
展开全部
设∠B=a,则∠C=2a,∠A=180-∠B-∠C=180-3a
AD是角平分线
∠BAD=∠DAC=∠A/2=90-3a/2
∠ADC=180-∠DAC-∠C=180-(90-3a/2)-2a=90-a/2
AC/sinB=AB/sinC
AC=ABsinB/sinC=ABsina/sin2a=ABsina/(2sinacosa)=AB/(2cosa)
AC/sin∠ADC=DC/sin∠DAC
DC=ACsin∠DAC/sin∠ADC=ACsin(90-3a/2)/sin(90-a/2)
=AB/(2cosa)*cos(3a/2)/cos(a/2)
AC+DC=AB/(2cosa)+AB/(2cosa)*cos(3a/2)/cos(a/2)
=AB(1+cos(3a/2)/cos(a/2))/(2cosa)
=AB(1+(4cos(a/2)^3-3cos(a/2))/cos(a/2))/(2cosa)
=AB(1+4cos(a/2)^2-3)/(2cosa)
=AB(4cos(a/2)^2-2)/(2cosa)
=AB(2cos(a/2)^2-1)/cosa
=ABcosa/cosa
=AB
证毕
AD是角平分线
∠BAD=∠DAC=∠A/2=90-3a/2
∠ADC=180-∠DAC-∠C=180-(90-3a/2)-2a=90-a/2
AC/sinB=AB/sinC
AC=ABsinB/sinC=ABsina/sin2a=ABsina/(2sinacosa)=AB/(2cosa)
AC/sin∠ADC=DC/sin∠DAC
DC=ACsin∠DAC/sin∠ADC=ACsin(90-3a/2)/sin(90-a/2)
=AB/(2cosa)*cos(3a/2)/cos(a/2)
AC+DC=AB/(2cosa)+AB/(2cosa)*cos(3a/2)/cos(a/2)
=AB(1+cos(3a/2)/cos(a/2))/(2cosa)
=AB(1+(4cos(a/2)^3-3cos(a/2))/cos(a/2))/(2cosa)
=AB(1+4cos(a/2)^2-3)/(2cosa)
=AB(4cos(a/2)^2-2)/(2cosa)
=AB(2cos(a/2)^2-1)/cosa
=ABcosa/cosa
=AB
证毕
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
延长AC,至F,使CF=DC,连接DF
∵△CDF是等腰三角形
∴∠CDF=∠CFD
∴2∠CFD=∠ACD
∵2∠B=∠C
∴∠B=∠CFD
∵∠BAD=∠DAC,AD为共边
∴△ABD≌△ADF
∴AB=AF
∴AB=AC﹢DC
∵△CDF是等腰三角形
∴∠CDF=∠CFD
∴2∠CFD=∠ACD
∵2∠B=∠C
∴∠B=∠CFD
∵∠BAD=∠DAC,AD为共边
∴△ABD≌△ADF
∴AB=AF
∴AB=AC﹢DC
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
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