flyingxiaobing
幼苗
共回答了12个问题采纳率:75% 举报
![](https://img.yulucn.com/upload/7/47/747d00c3e0a9c07104a0cbfb1fb92c49_thumb.jpg)
(1)由题意得:b+3=2c-8=0,(1分)
∴b=-3,c=4.(2分)
∴B(-3,0),C(0,4).(3分)
(2)∵CD∥AB,
∴∠DCB+∠ABC=180°.
∵∠COB=90°,
∴∠CBO+∠BCO=90°.(4分)
∵(∠GCF+∠DCB+∠BCO)+(∠CBO+∠ABC+∠ABM)
=180°+180°=360°,
∴∠ABM+∠GCF=360°-180°-90°=90°.(5分)
又∵∠CMB=∠MEA-∠ABM=70°-∠ABM
∠CNB=∠GCF-∠CFB=∠GCF-30°(6分)
∴∠CMB-∠CNB=(70°-∠ABM)-(∠GCF-30°)
=100°-(∠ABM+∠GCF)
=100°-90°
=10°.
(3)答:①
∠DQB+∠QBC
∠QPC的值不变,定值为2.
![](https://img.yulucn.com/upload/0/33/03360eef3a7045cd3330d8145c294b5a_thumb.jpg)
∵CP平分∠DCB,
∴∠QCB=2∠PCB.
又∵∠DQB=∠QBC+∠QCB,
∴∠DQB+∠QBC
=(∠QBC+∠QCB)+∠QBC
=2∠QBC+2∠PCB
=2(∠QBC+∠PCB)
=2∠QPC
∴②
∠DQB+∠QBC
∠QPC=
2∠QPC
∠QPC=2.(12分)
1年前
7