BCD is a trapezium with AB || DC. Eand F are points on non-parallel si

1.	BCD is a trapezium with AB \|\| DC. Eand F are points on non-parallel sides AD and BCrespectively such that EF is parallel to ABAE BF(see Fig. 6.14). Show that EDC
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.

BCD is a trapezium with AB || DC. Eand F are points on non-parallel sides AD and BCrespectively such that EF is parallel to ABAE BF(see Fig. 6.14). Show that EDC

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.