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  • ABCD is a trapezium with . E and F are points on non-parallel sides AD and BC respectively, such that

ABCD is a trapezium with AB \parallel CD. E and F are points on non-parallel sides AD and BC respectively, such that EF \parallel AB. Show that  \frac{AE}{ED}=\frac{BF}{FC}.

 

 

 

 
 
 
 
 

Answers (1)

Let us join AC to intersect EF at G.

AB\parallel DC  and  EF\parallel AB  (given)

\therefore EF\parallel DC

{line parallel to the same line are parallel to each other} 

\text{In }\bigtriangleup ADC, EG\parallel DC

\frac{AE}{ED}=\frac{AG}{GC} \;\;\;\;\;\; - (1)  (By proportionality theorem)

Similarly,

\text{In }\bigtriangleup CAB, GF\parallel AB

\frac{CG}{GA}=\frac{CF}{FB}   (by basic proportionality)

\frac{AG}{GC}=\frac{BF}{FC} \;\;\;\;\;\;\;\; - (2)

From eq (1) and eq (2)

\frac{AE}{ED}=\frac{BF}{FC}

Hence proved.

Posted by

Ravindra Pindel

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