如图,直线AB、CD相交于点O,OE⊥CD,OF⊥AB,图中与∠2互补的角共有( )A.0个B.1个C.2个D.3个
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如图,直线AB、CD相交于点O,OE⊥CD,OF⊥AB,图中与∠2互补的角共有( )
![](http://img.shitiku.com.cn/uploads/allimg/20191024/20191024090002-42491.png) |
答案
∵OE⊥CD,OF⊥AB, ∴∠2+∠EOB=90°,∠EOB+∠BOD=90°, ∴∠2=∠BOD=∠1, ∵∠BOD+∠COB=180°,∠BOC=∠AOD, ∴∠2+∠COB=180°,∠2+∠AOD=180°, ∴与∠2互补的角只有∠COB与∠AOD两个. 故选:C. |
举一反三
如图,直线AB、CD交于点O, (1)若∠AOC=90°,则AB______CD. (2)若AB⊥CD,则∠AOC=______=______=______=______度.
![](http://img.shitiku.com.cn/uploads/allimg/20191024/20191024085957-56550.png) |
如图,AD⊥BD,BC⊥CD,AB=acm,BC=bcm,则BD的取值范围是( )A.小于bcm | B.大于acm | C.大于acm或小于bcm | D.大于bcm且小于acm |
![](http://img.shitiku.com.cn/uploads/allimg/20191024/20191024085954-53055.png) |
如图,直线AB、CD、EF都经过点O,AB⊥CD,∠AOE=70°,求∠BOF、∠DOF的度数.
![](http://img.shitiku.com.cn/uploads/allimg/20191024/20191024085950-85313.png) |
已知:四图,AO⊥BO,∠1=∠3.求证:上O⊥DO.
![](http://img.shitiku.com.cn/uploads/allimg/20191024/20191024085945-31750.png) |
∠A与∠B的两边分别垂直,试画图并猜想∠A与∠B数量关系. |
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