Any counterexample to answer this question on elementary geometry? Question: See the figure below. If AB=BC and DE=EF, is the line DF parallel to the line AC?

This should be an elementary problem. But I can't construct a counterexample to disprove the above question. If the answer is negative, please give a counterexample. Thanks.
 A: Hint:


*

*Let $D'F'$ be a segment analogous to $DF$ such that $E \in D'F'$ and $D'F' \parallel AC$.

*Then $\triangle DED'$ and $\triangle FEF'$ are congruent and so $DD' \parallel FF'$.


I hope this helps $\ddot\smile$
A: You can think your problem in terms of projective geometry: let $r$ be the line containing $D,E,F$ with $R$ being its point at the infinity. Similarly let $s$ the line containing $A,B,C$ with $S$ being its point at the infinity.Finally, let $P$ be the not yet labelled vertex of your triangle. 
Consider the projectivity $f:r\rightarrow s$ given by projection (to $P$) and section (with $s$), you have that $f(D)=A$, $f(E)=B$ and $f(F)=C$. This gives an equality of cross-ratios $$(D,E;F,R)=(A,B;C,f(R)).$$
Your condition is that $(D,E;F,R)=(A,B;C,S)=2$ (the distance from $D$ to $F$ is twice the distance from $E$ to $F$ and similarly for $A,B,C$). Anyhow, you have the equality $(A,B;C,f(R))=(A,B;C,S)$, which implies that $f(R)=S$.
But now, $f(R)$ is just the point where the line through $P$ and $R$ cuts the line $s$. Since this point is $S$, belonging to the infinity line, you must have $R=S$, meaning that $r$ and $s$ cut at the infinity and hence they are parallel.
N.B.: Note that I didn't use that $DE=EF$ and $AB=BC$, but only that $DF/EF=AC/BC$ (it doesn't matter if the quotient equals 2, like in your situation, or any other number).
A: This comes from the Intercept Theorem, also known as Thales Theorem. It states that if two line that intersect each other (aren't parallel) at point $S$, cut two other parallel lines such that the first lines cuts the parallel lines at points $A,B$ and the second line cuts the parallel lines at $C,D$ then the ratios of any two segments on the first line equals the ratios of the according segments on the second line.
Also this holds other way around.
So from $AB=BC$ and $DE=DF$ we have:
$$\frac{AB}{BC} = \frac{DE}{DF}$$
Also we know that the lines through points $AD$ adn $FC$ intersect at one point. So it follows that $$AC\mid \mid DF$$
