如图,AB‖CD,∠DAB=∠BCD,AE、BE分别平分∠DAB、∠ABC,求∠E
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如图,AB‖CD,∠DAB=∠BCD,AE、BE分别平分∠DAB、∠ABC,求∠E
![](http://img.wesiedu.com/upload/8/f0/8f0b8a327e1cbac60b42fb7522c49b67.jpg)
![](http://img.wesiedu.com/upload/8/f0/8f0b8a327e1cbac60b42fb7522c49b67.jpg)
![如图,AB‖CD,∠DAB=∠BCD,AE、BE分别平分∠DAB、∠ABC,求∠E](/uploads/image/z/17252395-43-5.jpg?t=%E5%A6%82%E5%9B%BE%2CAB%E2%80%96CD%2C%E2%88%A0DAB%3D%E2%88%A0BCD%2CAE%E3%80%81BE%E5%88%86%E5%88%AB%E5%B9%B3%E5%88%86%E2%88%A0DAB%E3%80%81%E2%88%A0ABC%2C%E6%B1%82%E2%88%A0E)
∵AB∥CD
∴∠ABC+∠BCD=180°
∵∠DAB=∠BCD
∴∠ABC+∠DAB=180°
∵AE、BE分别平分∠DAB、∠ABC
∴∠EAB=½∠DAB
∠EAB=½∠ABC
∴∠E=180°-½(∠DAB+∠ABC)=90°
望采纳,不清楚的地方可追问我
再问: 能帮我把理由标一下么?
再答: ∵AB∥CD(理由) ∴∠ABC+∠BCD=180° ∵∠DAB=∠BCD(理由) ∴∠ABC+∠DAB=180° ∵AE、BE分别平分∠DAB、∠ABC(理由) ∴∠EAB=½∠DAB ∠EAB=½∠ABC ∴∠E=180°-½(∠DAB+∠ABC)=90°
∴∠ABC+∠BCD=180°
∵∠DAB=∠BCD
∴∠ABC+∠DAB=180°
∵AE、BE分别平分∠DAB、∠ABC
∴∠EAB=½∠DAB
∠EAB=½∠ABC
∴∠E=180°-½(∠DAB+∠ABC)=90°
望采纳,不清楚的地方可追问我
再问: 能帮我把理由标一下么?
再答: ∵AB∥CD(理由) ∴∠ABC+∠BCD=180° ∵∠DAB=∠BCD(理由) ∴∠ABC+∠DAB=180° ∵AE、BE分别平分∠DAB、∠ABC(理由) ∴∠EAB=½∠DAB ∠EAB=½∠ABC ∴∠E=180°-½(∠DAB+∠ABC)=90°
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