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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 then
Option: 1
$\frac{ AE }{ ED }=\frac{ BF }{ FC }$

Option: 2
$\frac{ AD }{ ED }=\frac{ BC }{ FC }$

Option: 3
$\frac{ AD }{ AE }=\frac{ BC }{ BF }$

Option: 4
All of the above

Answers (1)

best_answer

Let us join AC to intersect EF at G

AB II DC and EF II AB (Given)

So, EF || DC (Lines parallel to the same line are parallel to each other

Now , in triangle ADC

EG II DC (As EF || DC )

So, $\frac{A E}{E D}=\frac{A G}{G C} \quad\text{ (Theorem 1)}$

Similarly, from ? CAB,

$\\\frac{ CG }{ AG }=\frac{ CF }{ BF } \\ \\\frac{ AG }{ GC }=\frac{ BF }{ FC }$

Therefore, $\frac{A E}{E D}=\frac{B F}{F C}$

On adding 1 both side

$\frac{A E}{E D}+1=\frac{B F}{F C}+1$

$\frac{A E}{E D}+\frac{ED}{E D}=\frac{B F}{F C}+\frac{FC}{FC}$

$\frac{A D}{E D}=\frac{B C}{F C}$

$\frac{A E}{E D}=\frac{B F}{F C}\Rightarrow\frac{ED}{AE}=\frac{FC}{BF}$

on adding 1 both side

$\frac{E D}{A E}+\frac{AE}{AE}=\frac{F C}{B F}+\frac{B F}{B F}$

$\frac{A D}{AE}=\frac{B C}{B F}$

Hence all options are true

Posted by

Irshad Anwar

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