Question

求达人帮忙看看，这道题该怎么做，要有过程哦，谢谢

Answer 1

解：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、EG三等分∠AEC
∴∠AEF＝∠AEC/3＝30°
∴∠AEF＝∠BAE

∴EF∥AB

Answer 2

解:(1)<AEF=30 连接ED ∵AB∥Dc ∴<ADC=<BAC=30 ∵<DCE=60 ∴在△EcD中,<cED=90 ∵EF，EG三等平分<AEc ∴<AEF二90÷3=30 (2)证明:由(1)可知<AEF=30∵<BAE=30∴<BAE=<AEF∴AB∥EF

Answer 3