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如图,已知四边形ABCD中,E,F分别在AB,CD上,且AB/AE=DC/DF,求证(1)AB/EB=DC/FC(2)(AB+DC)/(EB+FC)=(AB-DC)/(EB-FC)
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1、AB/AE=DC/DF
那么AB/(AB-AE)=dc/(dc-df)
即AB/EB=DC/FC
2、AB/EB=DC/FC
∴AB/DC=EB/FC
那么(AB+DC)/DC=(EB+FC)/FC……(1)
(AB-DC)/DC=(EB-FC)/FC……(2)
两式相除
(AB+DC)/(AB-DC)=(EB+FC)/(EB-FC)
(AB+DC)/(EB+FC)=(AB-DC)/(EB-FC)
那么AB/(AB-AE)=dc/(dc-df)
即AB/EB=DC/FC
2、AB/EB=DC/FC
∴AB/DC=EB/FC
那么(AB+DC)/DC=(EB+FC)/FC……(1)
(AB-DC)/DC=(EB-FC)/FC……(2)
两式相除
(AB+DC)/(AB-DC)=(EB+FC)/(EB-FC)
(AB+DC)/(EB+FC)=(AB-DC)/(EB-FC)
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