Question

在五边形ABCDE中,如图AB=AE,BC+DE=CD,∠ABC+∠AED=180°,试说明AD平分∠CDE的理由？

Answer 1

将△ADE以点A为中心,顺时针旋转与∠BAE相等的度数,到△AFB位置,连结DF.由△ADE≌△AFB,可得∠AED=∠ABF,∠ADE=∠AFB,ED=BF,AD=AF.因为∠ABC+∠AED=180°,所以∠ABC+∠ABF=180°,所以C、B、F三点共线.又因为CD=BC+DE=BC+BF=CF,所以∠CFD=∠CDF.由AD=AF,可得∠DFA=∠FDA,所以∠ADE=∠AFB=∠CFD+∠DFA=∠CDF+∠FDA=∠ADC.,9,