In trapezium ABCD, AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF || AB. Show that AE/BF = ED/FC.
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Answered by
164
Let us join AC to intersect EF at G .
AB // DC and EF // AB ( given )
=> EF // DC
[ Lines parallel to the same line are
parallel to each other. ]
In ∆ADC , EG // DC
AE/ED = AG/GC ---- ( 1 )
[ By Basic Proportionality theorem ]
Similarly ,
In ∆CAB , GF // AB
CG/GA = CF/FB
[ By basic Proportionality theorem ]
AG/GC = BF/FC ---( 2 )
from ( 1 ) and ( 2 ) ,
AE/ED = BF/FC
=> AE/BF = ED/FC
Hence proved.
I hope this helps you.
: )
AB // DC and EF // AB ( given )
=> EF // DC
[ Lines parallel to the same line are
parallel to each other. ]
In ∆ADC , EG // DC
AE/ED = AG/GC ---- ( 1 )
[ By Basic Proportionality theorem ]
Similarly ,
In ∆CAB , GF // AB
CG/GA = CF/FB
[ By basic Proportionality theorem ]
AG/GC = BF/FC ---( 2 )
from ( 1 ) and ( 2 ) ,
AE/ED = BF/FC
=> AE/BF = ED/FC
Hence proved.
I hope this helps you.
: )
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