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

(N/A) Let us join $AC$ to intersect $EF$ at $G$ (see Figure).
Given: $AB \parallel DC$ and $EF \parallel AB$.
Since lines parallel to the same line are parallel to each other,we have $EF \parallel DC$.
Now,in $\Delta ADC$,since $EG \parallel DC$ (as $EF \parallel DC$),by Basic Proportionality Theorem,we have:
$\frac{AE}{ED} = \frac{AG}{GC} \quad ...(1)$
Similarly,in $\Delta CAB$,since $GF \parallel AB$ (as $EF \parallel AB$),by Basic Proportionality Theorem,we have:
$\frac{CG}{AG} = \frac{CF}{BF}$
Taking the reciprocal of both sides,we get:
$\frac{AG}{GC} = \frac{BF}{FC} \quad ...(2)$
From equations $(1)$ and $(2)$,we get:
$\frac{AE}{ED} = \frac{BF}{FC}$.