Example 2 : ABCD is a trapezium with AB || DC.
E and F are points on non-parallel sides AD and BC
respectively such that EF is parallel to AB
AE FB
(see Fig. 6.14). Show that
ED CF
Solution : Let us join AC to intersect EF at G
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Answer:
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.
Explanation:
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