If a quadrilateral is a parallelogram, the diagonals form two congruent triangles - How can I prove this

Got a math problem - in parallelogram A-B-C-D, the diagonals bisect each other at point M, I have to prove that the triangles ABC and CDA are congruent. Just stating that they are congruent is not enough, and claiming SSS, ASA etc is apparently not good enough either.

I also have to prove that:
1. AB = DC (the parallel top and bottom sides)
2. AM = CM and BM = DM (the bisected diagonals)

Apart from stating quadrilateral theory for parallelograms, is there any way to PROVE all of these?
Thanks.

1) To Prove ΔABC ≅ ΔCDA

i) Join the diagonal AC.
ii) In triangles, ABC and CDA,


Similarly
AC = CA [Common side]

Hence, Δ ABC ≅ ΔCDA [ASA, that is, Angle-Side-Angle congruence axiom]

2) From the above, AB = CD [If two triangles are congruent, then their corresponding sides are equal; the corresponding sides are - (AB, BC, CA) & (CD, DA & AC)]

3) Join BD; so the two diagonals AC & BD intersect at M.
Considering the triangles, AMB & CMD,
As in (1) above,
AB = CD [From (2) above]

==> ΔAMB ≅ ΔCMD [ASA congruence axiom]

So, AM = CM and BM = DM [Corresponding parts of congruence triangles are equal]

Sure you can prove it. The quadrilateral theorem for parallelograms (I assume that's basically what all this states) isn't an axiom, is it? Go back to basic principles, applied to the a couple of pairs of parallel lines, and all this falls out.

angle BAC = angle ACD (alternate interior angles formed by transveral across 2 parallel lines)