CD ,C'D'分别是三角形ABC和A'B'C'的角平分线,CD=C'D',AB=A'B',角ADC=角A'D'C',求证这两个三角形全等

反正，设AC>A'C'
AC/sinADC=CD/sinA=AD/sinADC
另一三角形式子相同

AC>A'C'
CD=C'D',角ADC=角A'D'C'
A<A'
A+ADC+ACD=180
ACD>A'C'D'
所以AD>A'D'

ACB=ACD>A'C'D'=A'C'B'
B<B'
同样方法CD/sinB=BD/sinBDC
BD>B'D'
AB>A'B'
与题目不符
所以AC不大于A'C'，同理也不小于
所以AC=A'C'
同理BC=B'C'
够了 楼上请注意题目

证明

过点C做CE‖于AB,过点A做AE‖CD交CE于E,同样做过点C'做C'E'‖于A'B',过点A'做A'E'‖C'D'交C'E'于E'.

易得,四边形ADCE和A'D'C'E'是平行四边形→∠CDA=∠AEC,∠C'D'A'=∠A'E'C'
且CD=AE,C'D'=A'E'又∠ADC=∠A'D'C',CD=C'D'→∠AEC=∠A'E'C',AE=A'E',又∵AB=A'B',∴△AEB≌△A'E'B'→∠ECA=∠E'C'A',∠EAC=∠E'A'C'

∵四边形ADCE和A'D'C'E&#