已知三角形ABC,AB=AC,角A=20度,D在AB边上,AD=BC,连接DC,求证：角BDC=30度

题目

已知三角形ABC,AB=AC,角A=20度,D在AB边上,AD=BC,连接DC,求证：角BDC=30度

答案

设,角BDC=x
DC/sin80=BC/sinx.(1)
DC/sin20=AD/sin(x-20)=BC/sin(x-20).(2)
(1)/(2)
sin(x-20)/sinx=sin20/sin80=2sin10cos10/cos10=2sin10
sin(x-20)=2sin10sinx
sinxcos20-cosxsin20=2sinxsin10
sinx(cos20-2sin10)=cosxsin20
tanx=sin20/(cos20-2sin10)=sin20/(sin70-sin10-sin10)
=sin20/(2sin30cos40-sin10)
=sin20/(sin50-sin10)=sin20/2sin20cos30=根号3/3
x=30度
即角BDC=30度