五边形ABCDE中BC+DE=CD,AB=AE,角ABC+角AED=180度,求证:AD平分角CDE

题目

五边形ABCDE中BC+DE=CD,AB=AE,角ABC+角AED=180度,求证:AD平分角CDE

答案

连接AD
AB=AE -> △ADE绕A点旋转,使E点与B点重合,D点转至D'
-> AD=AD',BD'=DE,∠D'BA=∠AED,∠BD'A=∠ADE
∠ABC+∠AED=180°
-> ∠ABC+∠D'BA=∠ABC+∠AED=180°
-> D',B,C三点在同一直线上
BC+DE=CD,BD'=DE
-> CD'=BC+BD'=BC+DE=CD
连接DD',CD'=CD
-> ∠CD'D=∠CDD'
AD=AD'
-> ∠AD'D=∠ADD'
∠BD'A=∠CD'D+∠AD'D
∠ADC=∠CDD'+∠ADD'
-> ∠BD'A=∠ADC
∠BD'A=∠ADE
-> ∠ADC=∠ADE
-> AD平分角CDE