How to find $\frac{AC}{CE}-\frac{BD}{DF}$ given $\frac{AC}{AE}+\frac{DF}{BF}=1$? The problem is as follows:

Let $\textrm{A, B, C, D, E and F}$ collinear points and consecutive.
It is known that,
$$\frac{AC}{AE}+\frac{DF}{BF}=1$$
Find the value of,
$$\frac{AC}{CE}-\frac{BD}{DF}$$

The choices given are:
$\begin{array}{cc}
1.&-1\\
2.&1\\
3.&0\\
4.&2\\
5.&3\\
\end{array}$
The official answers sheet indicates that the answer is 0. But how do I get there?.
I've attempted assigning unknown for the segments in between those points like a, b, c, d, e. But by doing so this becomes into a maze of equations that doesn't seem to get me anywhere.
Please it would help if you don't try to force the solution. Can it be solved disregarding knowing beforehand the given choices and the answer?
Does it exist some shortcut or workaround for this problem?. Can someone please help me here? I am stuck.
 A: We have $\frac{AC}{AE} +\frac{DF}{BF} = 1$
$$=>\frac{AC}{AC +CE} + \frac{DF}{DF + BD} = 1$$
$$=>\frac{AC}{AC+CE}=1-\frac{DF}{DF+BD}$$
$$=>\frac{AC}{AC+CE}=\frac{BD}{DF+BD}$$
Now, If we take the reciprocal, We get: $\frac{AC+CE}{AC} = \frac{DF + BD}{BD}$
$$=>\frac{CE}{AC}= \frac{DF}{BD}$$
If we again take the reciprocal, We get: $\frac{AC}{CE} = \frac{BD}{DF}$
$$=>\frac{AC}{CE} -\frac{BD}{DF} = 0$$
A: You are given that:
$$\frac{AC}{AE} + \frac{DF}{BF} = 1$$
Let's split the denominators into sub-segments.
$$\frac{AC}{AC + CE} + \frac{DF}{BD + DF} = 1$$
Let $x = \frac{AC}{CE}$ and $y = \frac{BD}{DF}$.  We ultimately want to find $x - y$.  For now, make the substitutions $AC = (CE)x$ and $BD = (DF)y$.
$$\frac{(CE)x}{(CE)x + CE} + \frac{DF}{(DF)y+ DF} = 1$$
Reduce the fractions.
$$\frac{x}{x + 1} + \frac{1}{y + 1} = 1$$
Multiply everything by $(x + 1)(y + 1)$ in order to get rid of the fractions.
$$x(y+1) + x + 1 = (x + 1)(y + 1)$$
$$xy + 2x + 1 = xy + x + y + 1$$
$$x = y$$
So $x - y = 0$, Q.E.D.
