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解答 解:(1)∵∠MON=90°,∠NAB=∠AOB+∠ABO=90°+∠ABO,∠ABM=∠AOB+∠BAO=90°+∠BAO,∴∠NAB+∠MBA=90°+∠ABO+90°+∠BAO=90°+180°=270°,∵AD平分∠BAN,BC平分∠ABM,∴∠CAB+∠CBA=12×270°12×270°=135°,∴∠ACB=45°;(2)∠ACB的度数不改变如图1,∵AD平分∠BAN,BC平分∠ABM...
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(1)①∵∠MON=40°,OE平分∠MON∴∠AOB=∠BON=20°∵AB∥ON∴∠ABO=20°②∵∠BAD=∠ABD∴∠BAD=20°∵∠AOB+∠ABO+∠OAB=180°∴∠OAC=120°∵∠BAD=∠BDA,∠ABO=20°∴∠BAD=80°∵∠AOB+∠ABO+∠OAB=180°∴∠OAC=60°故答案为:①20 ②120,60(2)①当点D在线段OB上时,若∠BAD=∠ABD...
[题目]已知:∠MON=36°.OE平分∠MON.点A.B分别是射线OM.OE.上的动点.点D是线段OB上的动点.连接AD并延长交射线ON于点C.设∠OAC=x.(1)如图1.若AB∥ON.则①∠ABO的度数是 ,②当∠BAD=∠ABD时.x= ,当∠BAD=∠BDA时.x= ,(2)如图2.若AB⊥OM.则是否存在这样的x的值.使得△ABD
[题目]已知:∠MON=40°.OE平分∠MON.点A.B.C分别是射线OM.OE.ON上的动点.连接AC交射线OE于点D.设∠OAC= °.(1)如图1.若AB//ON.则①∠ABO的度数是 ,②当∠BAD=∠ABD时. = ,③当∠BAD=∠BDA时. = .(2)如图2.若AB⊥OM.则是否存在这样的x的值.使得△ADB中有两个相等的
解答 解:(1)①∵∠MON=80°,OE平分∠MON.∴∠AOB=∠BON=40°,∵AB∥ON,∴∠ABO=40°故答案是:40°;②如答图1,∵∠MON=80°,且OE平分∠MON,∴∠1=∠2=40°,又∵AB∥ON,∴∠3=∠1=40°,∵∠BAD=∠ABD,∴∠BAD=40°∴∠4=80°,∴∠OAC=60°,即x=60°.(2)存在这样的x,①如答图2,...
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