如图,EB和DC相交于点A,CF、EF分别平分∠ACB、∠AED.若∠B=70°,∠D=40°,则∠F的度数为···········( )(A)55° (B)56° (C)57° (D)58°
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/16 22:21:17
如图,EB和DC相交于点A,CF、EF分别平分∠ACB、∠AED.若∠B=70°,∠D=40°,则∠F的度数为···········()(A)55°(B)56°(C)57°(D)58°如图,EB和DC
如图,EB和DC相交于点A,CF、EF分别平分∠ACB、∠AED.若∠B=70°,∠D=40°,则∠F的度数为···········( )(A)55° (B)56° (C)57° (D)58°
如图,EB和DC相交于点A,CF、EF分别平分∠ACB、∠AED.若∠B=70°,∠D=40°,则∠F的度数为···········( )
(A)55° (B)56° (C)57° (D)58°
如图,EB和DC相交于点A,CF、EF分别平分∠ACB、∠AED.若∠B=70°,∠D=40°,则∠F的度数为···········( )(A)55° (B)56° (C)57° (D)58°
A
∠F+1/2∠ACB=∠D+1/2∠AED
∠F=∠D+1/2(∠AED-∠ACB)
又∠AED+∠D=∠B+∠ACB
所以∠F=40+1/2(70-40)=55
简单啊