Proportionality and Thales Theorem let $k$ and $l$ be parallel lines. Let $A, B, C$ be points on $k$ and $A' B', C'$ be points on $l$ such that the lines $AA', BB'$ and $CC'$ intersect at a common point. Prove that $\frac{|A'B'|}{|B'C'|}=\frac{|AB|}{|BC|}$.
I tried using Thales's theorem on proportionality but I got confused
Any hints,
Thanks
 A: If $O$ is the common point of $AA'$, $BB'$ and $CC'$, notice that triangle $ABO$ is similar to $A'B'O$ and triangle $BCO$ is similar to $B'C'O$. From that the assertion follows.
A: I think that you have to have two considerations, and they are, where is the point $P$ in which they intersect( inside or outside the parallels). I think (I am not completely sure) it depends on how your order the points. I´ll just give you one consideration, and it´s that the point $P$ is inside the parallels.

Here I just put five points ($A,C,A',C',P$),I know I haven´t put the other two points, but I can already say, that´s the point $P$, where they intersect. I can say that some angles, for example (angle $CAA'\cong AA'C'$, because they are angles between parallels) are equal to others. I´d like to say the name exact of them but, since I am not an english speaker I don´t know what they are(I´d like you to edit it, and put them). And now I can say that triangles $ACP \sim A'C'P$ by $AAA$. 

So now drawing this second graphic I get some pairs of angles that are equal (angle that measure the same have the same colors)because they are angles between lines (Opossed to the vertex? is that the correct word? and also I get other similar triangles. And because they are similar, I can establish the following proportions:
$1)$ $$\frac{|AB|}{|A'B'|}=\frac{|BP|}{|B'P|}$$
$2)$ $$\frac{|BC|}{|B'C'|}=\frac{|BP|}{|B'P|}$$
Therefore
$$\frac{|AB|}{|A'B'|}=\frac{|BC|}{|B'C'|}=\frac{|BP|}{|B'P|}$$
So, we have:
$$\frac{|AB|}{|A'B'|}=\frac{|BC|}{|B'C'|}$$
And then if I multiply by $|A'B'|$ and $\frac{1}{|BC|}$, I get to:
$$\frac{|AB|}{|BC|}=\frac{|A'B'|}{|B'C'|}$$ which what you are asking for.
But whoever asked this question could say to you, that was not what they meant, because when they asked the question, they didn´t specify the order of the points. So here would be there other solution (Of course, it requires a new graphic):
(The difference is the order of the points)
So, here we also have similar triangles, which, technically, are the same as before.
So, for example, $ACP \sim PA'C'$ by $AA$. So, I can establish the following proportionS:
$1)$ $$\frac{|AC|}{|A'C'|}=\frac{|CP|}{|C'P|}$$
$2)$ $$\frac{|CB|}{|C'B'|}=\frac{|CP|}{|C'P|}$$
Therefore:
$$\frac{|AC|}{|A'C'|}=\frac{|CB|}{|C'B'|}$$
(I am going to solve this problem using a property of proportions that I´ll also demonstrate, but you could solve it geometrically.)

So, I´ll aply what I´ve just demonstrated:
$$\frac{|AC|+|CB|}{|A'C'|+|C'B'|}=\frac{|AC|}{|A'C'|}=\frac{|CB|}{|C'B'|}$$
$$\frac{|AB|}{|A'B'|}=\frac{|CB|}{|C'B'|}$$
And we get to:
$$\frac{|A'B'|}{|C'B'|}=\frac{|AB|}{|CB|}$$. And, I´d say, in this case, they are the two solutions. I haven´t done the cases where, $P$, where they intersect are outside the parallels, but I think (I am not completely sure), there should be two solutions as well (changing the order). =)
