如图①所示,已知,BC∥OA,∠B=∠A=100°,试回答下列问题:⑴试说明:OB∥AC;⑵如图②,若点E、F在BC
如图①所示,已知,BC∥OA,∠B=∠A=100°,试回答下列问题:⑴试说明:OB∥AC;⑵如图②,若点E、F在BC上,且∠FOC=∠AOC,OE平分∠BOF.试求∠EO...
如图①所示,已知,BC∥OA,∠B=∠A=100°,试回答下列问题:⑴试说明:OB∥AC;⑵如图②,若点E、F在BC上,且∠FOC=∠AOC ,OE平分∠BOF.试求∠EOC的度数;⑶在⑵的条件下,若左右平行移动AC,如图③,那么∠OCB:∠OFB的比值是否随之发生变化?若变化,试说明理由;若不变,求出这个比值;⑷在⑶的条件下,当∠OEB=∠OCA时,试求∠OCA的度数.
展开
爆死UELQ
推荐于2016-05-20
·
TA获得超过139个赞
关注
(1)理由见解析;(2)40°;(3)不变,1:2;(4)60. |
试题分析:(1)由同旁内角互补,两直线平行证明. (2)由∠FOC=∠AOC,并且OE平分∠BOF得到∠EOC=∠EOF+∠FOC= ![](https://iknow-pic.cdn.bcebos.com/9c16fdfaaf51f3de0d9d59d597eef01f3a297950?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) (∠BOF+∠FOA)= ![](https://iknow-pic.cdn.bcebos.com/9c16fdfaaf51f3de0d9d59d597eef01f3a297950?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ∠BOA,算出结果. (3)先得出结论:∠OCB:∠OFB的值不发生变化,理由为:由BC与AO平行,得到一对内错角相等,由∠FOC=∠AOC,等量代换得到一对角相等,再利用外角性质等量代换即可得证; (4)由(2)(3)的结论可得. (1)∵BC∥OA, ∴∠B+∠O=180°,又∵∠B=∠A, ∴∠A+∠O=180°, ∴OB∥AC; (2)∵∠B+∠BOA=180°,∠B=100°, ∴∠BOA=80°, ∵OE平分∠BOF, ∴∠BOE=∠EOF,又∵∠FOC=∠AOC, ∴∠EOF+∠FOC= ![](https://iknow-pic.cdn.bcebos.com/9c16fdfaaf51f3de0d9d59d597eef01f3a297950?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) (∠BOF+∠FOA)= ![](https://iknow-pic.cdn.bcebos.com/9c16fdfaaf51f3de0d9d59d597eef01f3a297950?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ∠BOA=40°; (3)结论:∠OCB:∠OFB的值不发生变化.理由为: ∵BC∥OA, ∴∠FCO=∠COA, 又∵∠FOC=∠AOC, ∴∠FOC=∠FCO, ∴∠OFB=∠FOC+∠FCO=2∠OCB, ∴∠OCB:∠OFB=1:2; (4)由(1)知:OB∥AC, 则∠OCA=∠BOC, 由(2)可以设:∠BOE=∠EOF=α,∠FOC=∠COA=β, 则∠OCA=∠BOC=2α+β, ∠OEB=∠EOC+∠ECO=α+β+β=α+2β, ∵∠OEC=∠OCA, ∴2α+β=α+2β, ∴α=β, ∵∠AOB=80°, ∴α=β=20°, ∴∠OCA=2α+β=40°+20°=60. |
收起
为你推荐: