Question

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Answer 1

QUESTION

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 ?

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.

Answer 2

Step-by-step explanation:

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