$E$ is the mid-point of the side $AD$ of the trapezium $ABCD$ with $AB \parallel DC$. $A$ line through $E$ drawn parallel to $AB$ intersects $BC$ at $F$. Show that $F$ is the mid-point of $BC$.

(N/A) Given: $A$ trapezium $ABCD$ in which $AB \parallel DC$ and $E$ is the mid-point of the side $AD$. Also,$EF \parallel AB$.
To prove: $F$ is the mid-point of $BC$.
Construction: Join $AC$ which intersects $EF$ at $O$.
Proof: In $\triangle ADC$,$E$ is the mid-point of $AD$ and $EF \parallel DC$.
$[\because EF \parallel AB \text{ and } DC \parallel AB \Rightarrow AB \parallel EF \parallel DC]$
$\therefore O$ is the mid-point of $AC$. [Converse of mid-point theorem]
Now,in $\triangle CAB$,$O$ is the mid-point of $AC$ and $OF \parallel AB$.
$\Rightarrow OF$ bisects $BC$ [Converse of mid-point theorem].
Or $F$ is the mid-point of $BC$.
Hence,proved.