Question

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

Answer 1

Let us join AC to intersect EF at G

Given that,

$AB\parallel DCandEF\parallel AB$

$\Rightarrow EF\parallel DC$

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

Now, in $\Delta ADC$,

$EG\parallel DC$ [$\because EF\parallel DC$]

$\Rightarrow \frac{AE}{ED}=\frac{AG}{GC}.....\left(1\right)$ [Basic proportionality Theorem]

Similarly, from $\Delta CAB$,

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

$\Rightarrow \frac{AG}{GC}=\frac{BF}{FC}.......\left(2\right)$

Therefore, from (1) and (2),

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