ABCD is a trapezium with $A B ∥ D C$ . E and F are points on non-parallel sides AD and BC respectively such that $E F ∥ A B$ . Then, $\frac{A E}{E D} = \frac{B F}{F C}$ .

True

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False

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Solution

The correct option is A
True

Let us join AC to intersect EF at G

Given that,

$A B ∥ D C a n d E F ∥ A B$

$\Rightarrow E F ∥ D C$

[Lines parallel to the same line are parallel to each other]

Now, in $Δ A D C$ ,

$E G ∥ D C$ [ $∵ E F ∥ D C$ ]

$\Rightarrow \frac{A E}{E D} = \frac{A G}{G C} . . . . . (1)$ [Basic proportionality Theorem]

Similarly, from $Δ C A B$ ,

$\frac{C G}{A G} = \frac{C F}{B F}$

$\Rightarrow \frac{A G}{G C} = \frac{B F}{F C} . . . . . . . (2)$

Therefore, from (1) and (2),

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

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