Proving that the sides of a quadrilateral are parallel (neutral geometry) 
Let $\Box$ABCD be convex. Suppose that $\angle A \cong \angle C$ and $\angle B \cong \angle D$. Prove that $\overleftrightarrow{AB} \parallel
\overleftrightarrow{CD}$ and $\overleftrightarrow{AD} \parallel
\overleftrightarrow{BC}.$

I'm studying for my exam, and I'm having trouble proving this. With similar problems with different information, you would draw a diagonal and get a pair of congruent triangles. Then use Propositions 27 and 28. With this one, though, it doesn't seem like you can do that. Any help would be appreciated.
Here are some of the things we can assume in neutral geometry

*

*$\sigma (\Box ABCD) \leq 360$

*$\sigma (\triangle ABC) \leq 180$

*Euclid's Propositions 27 and 28

This is for a college class, by the way.

 A: we can make a proof by contradiction.
the general idea is that lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ meet at one point E to the far left (or to the far right) thus creating two triangles: $\triangle$BEC and $\triangle{AED}$.
we will use the converse of Euclid's fifth postulare to argue that angles $\angle{EBC}$ and $\angle{ECB}$ sum to less than $180°$ and so angles $\angle{EAD}$ and $\angle{EDA}$ sum to more than $180°$ because of the linear pair theorem, giving a contradiction.
proof:
assume that $\Box{ABCD}$ is not a parallelogram, then either lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ intersect or $\overleftrightarrow{BC}$ and $\overleftrightarrow{DA}$ intersect.
let's assume that $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ intersect and call that point $E$.
from the convexity of $\Box{ABCD}$ you can prove that $E$ does not lie neither on segment $\overline{AB}$ nor on segment $\overline{CD}$ ($\overline{AB}$ and $\overline{CD}$ are semiparallel).
so either $E\star A\star B$ ($A$ lies between $E$ and $B$) or $E\star B\star A$, we'll assume that $E\star A\star B$.
again from the convexity of $\Box$ABCD you can prove that $C$ lies between $E$ and $D$ ($\overline{AD}$ and $\overline{BC}$ are semiparallel).
$\overleftrightarrow{BC}$ is a transversal of $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ and they meet on the same side as $A$ of $\overleftrightarrow{BC}$, from the converse of Euclid's fifth postulate then $\mu(\angle{ABC}) + \mu(\angle{DCB}) < 180°$ ($\mu(\angle{ABC})$ is the measure of $\angle{ABC}$).
because $\angle{DAB} \cong \angle{DCB}$ and $\angle{ADC} \cong \angle{ABC}$, also $\mu(\angle{DAB}) + \mu(\angle{ADC}) < 180°$.
from the linear pair theorem, $\mu(\angle{CDA}) + \mu(\angle{EDA}) = 180°$ and $\mu(\angle{BAD}) + \mu(\angle{EAD}) = 180°$, this means that $\mu(\angle{EAD}) + \mu(\angle{EDA}) = 360° - (\mu(\angle{CDA}) + \mu(\angle{BAD})) > 360° - 180° = 180°$.
but this contradicts the Saccheri-Legendre theorem for the triangle $\triangle{ADE}$, meaning that our initial assumption was false and $\Box{ABCD}$ is a parallelogram.
if you want more clarifications on the details left behind feel free to ask.
