已知AB∥CD,∠AEC＝90°⑴当CE平分∠ACD时,求证:AE平分∠BAC

过E作直线EF平行于AB再答：则有角A=角AEF,又因为角AEC=角A+角C,所以角CEF=角C,所以EF平行于CD,所以AB平行于CD

证明：过点E作EF∥AB．∵EF∥AB，∴∠A=∠AEF；又∵∠AEC=∠A+∠C，∴∠AEC=∠AEF+∠C；而∠AEC=∠AEF+∠CEF，∴∠CEF=∠C，∴EF∥CD，∴AB∥CD．再问：不能

过E做∠AEC的角平分线EF,用同旁内角互补可得出AB//EF,EF//CD所以AB//CD

过点E作EH∥AB，∵AB∥CD，∴EH∥AB∥CD；∴∠AEH=∠BAE=40°，∠CEH=∠ECD=70°，∴∠AEC=∠AEH+∠CEH=110°；∵EF平分∠AEC，∴∠AEF=12∠AEC=

做EG平行AB则∠AEG=140°,∠CEG=110°所以∠AEC=360-140-110=110°又因为平分所以∠AEF=55°

证明：方法一.延长CE交AB于F,因为角AEC是三角形AEF的外角,所以角AEC=角A+角AFE,因为角AEC=角A+角C,所以角AFE=角C,所以AB//CD.（内错角相等,两直线平行）方法二.连结

过E点向左做一条直线EF使得EF//AB因为EF//AB所以∠BAE=∠AEF因为∠AEC=∠A+∠C=∠BAE+∠C=∠AEF+∠C又∠AEC=∠AEF+∠FEC所以∠FEC=∠C因为∠FEC和∠C

过AB的中点O作OF⊥CD于F,连接OC∵AB=AE+BE=8∴OA=1/2AB=4=OC∴OE=OA-AE=2∵∠CEB=30°,∠OFE=90°∴OF=1/2OE=1∴CF=√(OC²-

证明：过点E作FE‖AB,∴∠AEF=∠A,∵∠AEC=∠A+∠C,即∠AEF+∠CEF=∠A+∠C.∴∠CEF=∠C.∴EF‖CD,∴AB‖EF,CD‖EF.∴AB‖CD.

延长AE交CD于H∵AB∥CD∴∠AHC＝∠BAE＝30∴∠AEC＝∠AHC+∠DCE＝30+60＝90∵EF、EG三等分∠AEC∴∠GEF＝∠AEC/3＝90/3＝30证明：∵∠AEC＝90,EF、

证明：过E作EF∥AB，∵EF∥AB，∴∠A=∠AEF，又∵∠A+∠C=∠AEC，∴∠C=∠CEF，∴EF∥CD，又∵EF∥AB，∴AB∥CD．

连OE因为OA=OC,EA=EC所以三角形OECOEA全等（SSS)所以∠A=∠C所以∠CEO=∠AEO所以O在∠AEC的平分线上在△DCB和△DEA中{DB=DA{CB=EA{DC=DE∴△DCB≌

AE平分∠BAC理由：因为ab平行cd,所以∠bac+∠acd=180°又因为∠aec=90,所以∠ace+∠cae=90所以∠bac+∠acd=∠bae+∠cae+∠ace+∠dce即：∠bae+∠

证明：因为ab平行cd,所以∠bac+∠acd=180°又因为∠aec=90,所以∠ace+∠cae=90所以∠bac+∠acd=∠bae+∠cae+∠ace+∠dce即：∠bae+∠dce=90因为

过点e作ef||ab因为ab||cd所以ab||cd||ef则∠a+∠aef=180°（两条直线平行,同旁内角互补）∠c+∠cef=180°（两条直线平行,同旁内角互补）所以∠a+∠aec+∠c=∠a

过E作EF∥AB,∵EF∥AB,∴∠A=∠AEF（两直线平行内错角相等）,又∵AB∥CD,EF∥AB,∴EF∥CD,∴∠C=∠CEF（两直线平行内错角相等）,又∵∠AEC=∠AEF+∠CEF,∴∠AE

90°

图在哪.

过E做EF//AB∴∠AEF+∠A=180∴∠AEF=60∴∠FEC=120-60=60∴∠FEC+∠C=60+120+180∴EF//CD因为EF//AB所以AB//CD

过E点作EG//AB,点G在E左边,则∠AEG=∠A因为∠AEC=∠A+∠C,∠AEC=∠AEG+∠CEG,所以∠CEG=∠C,所以EG//CD又因为EG//AB,所以AB//CD