$l, m$ and $n$ are three parallel lines intersected by transversals $p$ and $q$ such that $l, m$ and $n$ cut off equal intercepts $AB$ and $BC$ on $p$ (see Fig). Show that $l, m$ and $n$ cut off equal intercepts $DE$ and $EF$ on $q$ also.

(N/A) We are given that $AB = BC$ and we have to prove that $DE = EF$.
Let us join $A$ to $F$ intersecting $m$ at $G$.
The quadrilateral $ADFC$ is divided into two triangles,namely $\Delta ACF$ and $\Delta AFD$.
In $\Delta ACF$,it is given that $B$ is the mid-point of $AC$ $(AB = BC)$ and $BG \parallel CF$ (since $m \parallel n$). By the converse of the mid-point theorem,$G$ is the mid-point of $AF$.
Now,in $\Delta AFD$,we have $G$ as the mid-point of $AF$ and $GE \parallel AD$ (since $m \parallel l$). By the converse of the mid-point theorem,$E$ is the mid-point of $DF$.
Therefore,$DE = EF$.
In other words,$l, m$ and $n$ cut off equal intercepts on $q$ also.