Proof:
Refer to the attached figure.
There are a couple of ways you can go at this. One is to show the sums of the marked angles are the same, hence ∠B ≅ ∠D. Instead, we're going to show that ΔABD ≅ ΔCDB, hence ∠A ≅ ∠C.
1. AB║DC and BC║AD . . . . given
2. BD is a transversal to both AB║DC and BC║AD . . . . given
3. ∠CBD ≅ ∠ADB . . . . alternate interior angles where a transversal crosses parallel lines are congruent
4. ∠CDB ≅ ∠ABD . . . . alternate interior angles where a transversal crosses parallel lines are congruent
5. BD ≅ BD . . . . reflexive property of congruence
6. ΔABD ≅ ΔCDB . . . . ASA postulate
7. ∠A ≅ ∠C . . . . CPCTC
∠A and ∠C are opposite angles of parallelogram ABCD, so we have shown what you want to have shown.
