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

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/07/13 13:17:39

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

在五边形ABCDE中,已知AB=AE BC+DE＝CD 角ABC+角AED＝180度连接AD 求证 AD平分角CDE 五边形ABCDE中BC+DE=CD,AB=AE,角ABC+角AED=180度,求证:AD平分角CDE 在五边形ABCDE中已知AB=AE,BC+DE=CD,角ABC+角AED=180 连接AD,求；AD平分角CDE 在五边形ABCDE中,AB=AE、BC+DE=CD,∠ABC+∠AED=180°. 求证：AD平分∠CDE. （初二）在五边形ABCDE中,AB=AE、BC+DE=CD,∠ABC+∠AED=180°.求证：AD平分∠CDE. 五边形ABCDE中,AB=AE,AC+DE=CD,∠ABC+∠AED=180度.连接AD,求证AD平分∠CDE 如图,五边形ABCDE中,AB=AE,BC+DE=CD,∠ABC+∠AED=180°,连接AC、AD,求证：AD评分∠C 五边形ABCDE中,AB=AE,BC=DE,点F是CD的中点,且AF平分CD,求证：∠ABC=∠AED 如图,在五边形ABCDE中,AB=AE,BC +DE=CD,∠ABC+∠AED=180°,连接AC、AD.试判断AD是否五边形ABCD中,AB=AE,BC+DE=CD,角BAE=角BCD=120度,角ABC+角AED=180度,连接AD,求如图五边形abcde中,角bae等于m度AB等于AE,BC+DE等于CD,角ABC+角AED等于180度五边形ABCD中,AB=AE,BC=DE,角ABC=角AED点F是CD中点.求证AF垂直CD