How to find $\frac{AC}{CE}-\frac{BD}{DF}$ given $\frac{AC}{AE}+\frac{DF}{BF}=1$?

Question

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.

How is zero even a possible answer, unless you have zero-length line segments? — Dan, May 25 '22 at 00:10
It probably helps to observe that $BD+DF=BF$ and $AC+CE=AE$. — Robert Shore, May 25 '22 at 00:11
I can find that $\frac{AC}{CE} = \frac{BD}{DF}$, but not that either of those fractions must be zero. — Dan, May 25 '22 at 20:45
There must be a typo in the expression to find: it should be ${AC\over CE}-{BD\over DF}$. — Intelligenti pauca, May 25 '22 at 22:07
@Dan The question has been updated. I am sorry there was a mistake when typing the question. Can you please share the solution? — Chris Steinbeck Bell, May 26 '22 at 01:31
@Intelligentipauca Yes you are right it was a typo. I consulted with the original source. I've updated the question to reflect the changes. With that change done, is it solveable?. Help please? — Chris Steinbeck Bell, May 26 '22 at 03:14

VVR · Accepted Answer · 2022-05-26T16:59:45.057

3

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: $\frac{AC}{CE} = \frac{BD}{DF}$

$$=>\frac{AC}{CE} -\frac{BD}{DF} = 0$$

edited May 26 '22 at 16:59

answered May 26 '22 at 07:25

VVR

479

score 0 · Answer 2 · answered May 26 '22 at 01:50

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.

How to find $\frac{AC}{CE}-\frac{BD}{DF}$ given $\frac{AC}{AE}+\frac{DF}{BF}=1$?

2 Answers2