证明题.如图,AB平行CD.AD,BC相交于点E,过点E作EF平行AB,交BD于点F.(1)求证:1/AB + 1/CD =1/EF.

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证明题.如图,AB平行CD.AD,BC相交于点E,过点E作EF平行AB,交BD于点F.(1)求证:1/AB + 1/CD =1/EF.

证明题.如图,AB平行CD.AD,BC相交于点E,过点E作EF平行AB,交BD于点F.(1)求证:1/AB + 1/CD =1/EF.
证明题.
如图,AB平行CD.AD,BC相交于点E,过点E作EF平行AB,交BD于点F.
(1)求证:1/AB + 1/CD =1/EF.

证明题.如图,AB平行CD.AD,BC相交于点E,过点E作EF平行AB,交BD于点F.(1)求证:1/AB + 1/CD =1/EF.
∵AB‖EF‖CD
∴△DEF∽△DAB
∴EF/AB=DF/DB ①
同理 △BEF∽△BCD
∴EF/CD=BF/BD=(BD-DF)/BD=1-DF/BD
∴DF/BD=1-EF/CD(上式变形得到) ②
结合①②,得1-EF/CD=EF/AB
∴EF/AB+EF/CD=1
两边同除EF,得
1/AB+1/CD=1/EF
∴1/AB + 1/CD =1/EF

EF‖AB
EF/AB =DF /BD(1)
EF ‖CD
EF/CD=BF/BD (2)
(1)+(2)
EF/AB +EF/CD=DF /BD+BF/BD
=(DF +BF )/BD=1
两边同时除以EF
1/AB + 1/CD =1/EF.