Proof of Prop. I.27 in Euclid's Elements 
Let´s suppose I have lines $\overleftrightarrow{EB}, \overleftrightarrow{CD}$ and $\overleftrightarrow{EF}$, as shown in the picture and , and angle $BCD$ is equal to angle $CEF$, how can I prove $\overleftrightarrow{CD}$ is parallel to $\overleftrightarrow{EF}$? I "made" one demonstration but I would like to know how you´d do it.
So, this is how I proved it.

$Demonstration:$
$Hypothesis:$ We have straight lines as show in the picture ... and angles $BCD\cong CEF$.
$Thesis:$ $\overleftrightarrow{CD}$ is parallel to $\overleftrightarrow{EF}.$
So, let´s suppose that they are not parallel, therefore, there is a point $P$, in which $\overleftrightarrow{CD}$ and $\overleftrightarrow{EF}$ intersect, so that I can make the triangle $CEP$.
Because it´s a triangle,  its interior angles sum to $180°$ degrees. Supposing that angle $CEF$ equals to $x$ degrees, I can say that angle $ECP$ is equal to $180 - x $, because both have to sum to $180°$. So now, supposing that triangle $CEP$ is a real triangle, angle $P$ has to measure $a°$ ($a$, being a positive real number and $a\neq 0$), so when I sum the angles, I get: $$ x + (180 - x) +a = 180
$$ $$ 180 + a = 180
$$ $$a=0
$$
So, I end up having a contradiction, thefore $\overleftrightarrow{CD}$ is parallel to $\overleftrightarrow{EF}$.
I could have also, as you say, proven it by contradiction, using the external angle theorem, because (I think) that would suggest, in this case angle $P=0$.
So what do you think? . I problably posted this, because I think I am being redudant, and also because it is not exactly how Euclid proves it.
@Aretino, I didn´t know how to make a comment and put a phot in it so I had to edit the comment. So I this is what I thought: 

So, you first follow the steps and you get that $\angle BEC \cong \angle FEA$. And by $LAL$ you have that $\triangle BEC \cong \triangle FEA$ and $\angle EBC \cong \angle FAE \cong \angle HAI$, and I think that´s how we would get to a contradiction.
 A: Your proof looks fine, but you are using a result (sum of the interior angles of a triangle) which requires the parallel postulate. This theorem can be proved without using that postulate (and in a simpler way) by the exterior angle theorem: in triangle $CEP$ (I'm referring to the construction shown in the question) exterior angle $\angle BCD$ must be greater than remote interior angle $\angle CEP$, but in our case those two angles are equal, so we get a contradiction.
Of course I'm referring to Euclid's Exterior Angle Theorem, not to the theorem unfortunately given the same name by some high-school books (see the Wikipedia page linked above for details).
EDIT.
As the above description is not clear enough, here's a more detailed version of the proof. We will use Euclid's Exterior Angle Theorem: 

An exterior angle of a triangle is greater than either
  of the interior angles not adjacent to it.

This is a fundamental result in absolute geometry because its proof does not depend upon the parallel postulate. For the proof, see the Wikipedia page linked above, or Euclid's Elements.
We can now proceed to prove the theorem you asked about: lines $CD$ and $EF$ make equal corresponding angles $\angle BCD$ and $\angle CEF$ with line $BC$; we must prove that $CD$ and $EF$ are parallel.
Suppose, by contradiction, that $CD$ and $EF$ are not parallel and thus meet at some point $P$. A triangle $PCE$ is then formed, but that triangle has an exterior angle ($\angle BCD$ in diagram below) which is equal to an interior angle not adjacent to it ($\angle CEF$). That is not possible by the exterior angle theorem, according to which $\angle BCD$ must be greater than $\angle CEF$. 
We thus get a contradiction and $CD$ is parallel to $EF$.

It is worth noting that this theorem allows one to construct a line parallel to a given line and passing through a given point: the existence of such a line can then be proven in absolute geometry, as it doesn't require the parallel postulate. The latter is needed to ensure the uniqueness of the constructed parallel line.
A: What version of the parallel postulate are you using?  Your problem clearly requires it.  The converse of what you are trying to prove is one of the common equivalents.  If your parallel postulate is that the angles of a triangle sum to $\pi$, you can imagine the lines meeting to the right and making a triangle.  You can show the angles sum to more than $\pi$.  Then do the same to the left.
