I think I have to prove that the other opposite sides are also parallel, but I am not sure how to do it. It is given that the other opposite sides are congruent and parallel.
The exact nature of what constitutes a proof in geometry can be somewhat elusive; it is often a function of those facts you are already assumed to know. The level of rigour that is required of you is something we as outsiders can only guess at, so I will limit my comments to somewhat casual ones.
First, let's fix notation. Let us call the quadrilateral ABCD as we proceed around the vertices. Let us further assume that AB and CD are the given parallel and congruent sides.
Now, draw the diagonal BD, and note that BD is (part of) a transversal to the the two parallel lines containing AB and CD. Therefore, the angles <ABD and <BDC are equal (why?); call their common measure θ. Now, the triangles ΔABD and ΔBDC are composed of the same side-angle-side configuration. In the first case, AB-θ-BD, in the second case, BD-θ-DC. If you already "know" that two triangles with the same S-A-S measures are congruent, then you are essentially done, since this implies that the remaining side in each triangle (AD, resp. BC) are in fact congruent, and further, that the corresponding angles <BAD and <BCD are equal. Hence, the two sides AD an BC must be parallel, and therefore, the quadrilateral is a parallelogram.
Hope this is clear.