已知AB‖EF‖CD,AC,BD交于E点,(1)试说明1/AB +1/CD =1/EF (2)若AB=6厘米,CD=12厘米,求EF的长如题
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 03:13:07
![已知AB‖EF‖CD,AC,BD交于E点,(1)试说明1/AB +1/CD =1/EF (2)若AB=6厘米,CD=12厘米,求EF的长如题](/uploads/image/z/13260704-32-4.jpg?t=%E5%B7%B2%E7%9F%A5AB%E2%80%96EF%E2%80%96CD%2CAC%2CBD%E4%BA%A4%E4%BA%8EE%E7%82%B9%2C%281%29%E8%AF%95%E8%AF%B4%E6%98%8E1%2FAB+%2B1%2FCD+%3D1%2FEF+%282%29%E8%8B%A5AB%3D6%E5%8E%98%E7%B1%B3%2CCD%3D12%E5%8E%98%E7%B1%B3%2C%E6%B1%82EF%E7%9A%84%E9%95%BF%E5%A6%82%E9%A2%98)
已知AB‖EF‖CD,AC,BD交于E点,(1)试说明1/AB +1/CD =1/EF (2)若AB=6厘米,CD=12厘米,求EF的长如题
已知AB‖EF‖CD,AC,BD交于E点,(1)试说明1/AB +1/CD =1/EF (2)若AB=6厘米,CD=12厘米,求EF的长
如题
已知AB‖EF‖CD,AC,BD交于E点,(1)试说明1/AB +1/CD =1/EF (2)若AB=6厘米,CD=12厘米,求EF的长如题
1.EF/AB=CF/CB EF/CD=BF/BC EF/AB+EF/CD=CF/CB+BF/BC=BC/BC=1
有1/AB +1/CD =1/EF
2.代人上式 1/6+1/12=1/EF=1/4 EF=4
(1)在△abc中,∵ef//ab,∴有fc/bc=ef/ab
在△bcd中,∵ef//cd,∴有bf/bc=ef/cd,bf=bc-fc,∴bf/bc=(bc-fc)/bc=1-(fc/bc)=ef/cd
∵fc/bc=ef/ab,∴1-(fc/bc)=1-ef/ab=ef/cd
∴ 1-ef/ab=ef/cd
∴ef/ef-ef/ab=ef/cd,约去ef
全部展开
(1)在△abc中,∵ef//ab,∴有fc/bc=ef/ab
在△bcd中,∵ef//cd,∴有bf/bc=ef/cd,bf=bc-fc,∴bf/bc=(bc-fc)/bc=1-(fc/bc)=ef/cd
∵fc/bc=ef/ab,∴1-(fc/bc)=1-ef/ab=ef/cd
∴ 1-ef/ab=ef/cd
∴ef/ef-ef/ab=ef/cd,约去ef
∴1/ef-1/ab=1/cd
∴1/ab+1/cd=1/ef
(2)把ab=6cm,cd=12cm代入方程1/ab+1/cd=1/ef
1/6+1/12=1/ef
∴ef=4cm
收起