Question

△ABC中,∠ABC=∠ACB,BD的延长线交外角∠ACM的平分线于E.直线CE与直线AB交于F △ABC中,∠ABC=

Answer 1

（1）1.因为BD平分∠ABC,∠ABC=∠ACB

所以∠DBC=1/2∠ABC

由三角形外角公式得：∠ADB=∠ADB+∠ACB=3/2∠ABC

由对顶角公式：∠CDE=∠ADB=3/2∠ABC=36

因为∠FCM=∠ABC+∠F 所以∠F=∠FCM-∠ABC

而CE平分∠ACM所以∠FCM=1/2(180-∠ACB)=90-1/2∠ABC

所以∠F=90-1/2∠ABC-∠ABC=90-3/2∠ABC=54

2.同1：∠CDE=3/2∠ABC=48

∠F=90-3/2∠ABC=48

3.由1、2得∠F+∠CDE=90

（2）结论：∠F+∠CDE=90

证明：由三角形外角公式和对顶角公式得：

∠CDE=∠ADB=∠DBC+∠ACB=1/2∠ABC+∠ABC=3/2∠ABC

∠F=∠FCM-∠ABC=1/2(180-∠ACB)-∠ABC=90-3/2∠ABC

所以∠F+∠CDE=90-3/2∠ABC+3/2∠ABC=90

即∠F+∠CDE=90 可得等式成立于∠BAC的大小无关