Quadrilaterals that has congruent opposite sides is parallelograms Quadrilaterals that has congruent opposite sides is parallelograms.
The following is a proof.
for quadrilateral ABCD,
AB = CD, BC = AD, AC = AC
hence ABC = CDA (SSS) 
mBAC = mDCA (alternate interior angle theorem) 
hence
AB // DC 
mACB = mCAD (alternate interior angle theorem)
AD // BC 
(Q.E.D)
but, I don't know why D, B is opposite by line AC for the alternate interior angle theorem.
How to prove it?
How to prove that ABCD is convex?
 A: In your proof the point that $\Delta ABC\cong\Delta CDA$ is right, but the following step is wrong, because you need to prove that $B$ and $D$ are placed in the different sides respect to $AC$, which is not so obvious.
Indeed, let $B$ and $D$ are placed in the same side respect to $AC$.
Since $\Delta ABC\cong\Delta CDB$, we obtain:
$$\measuredangle BAC=\measuredangle DCA$$ and
$$\measuredangle BCA=\measuredangle DAC.$$
Now, if $\measuredangle BAC=\measuredangle DAC$ so rays $AB$ and $AD$ they are the same ray and rays $CB$ and $CD$ they are the same ray, which gives $B\equiv D$, which is a contradiction.
Let $\measuredangle BAC>\measuredangle DAC$.
Thus, the ray $AD$ is placed between sides of the angle $BAC$, 
which says that the ray $AD$ intersects a side $BC$.
Let it happens in the point $K$.
By the same way we obtain that the ray $CB$ intersects a side $BC$ and let it happens in the point $L$.
Id est, lines $AD$ and $BC$ intersects in $\{K,L\}$, which says $K\equiv L$ and sides $BC$ and $AD$ have a common point, which is a contradiction again.
Thus, $B$ and $D$ are placed in the different sides respect to $AC$, which gives that $\angle BAC$ and $\angle DCA$ are alternate angles between $AB$ and $CD$ and secant $AC$, which says $AB||CD.$
By the same way we obtain $BC||AD$, which gives $ABCD$ is a parallelogram and we are done!
Actually.
This theorem is one of hardest theorems  at the beginning of geometry. 
