1. **State the problem:** Given quadrilateral ABCD with AB \parallel CD and AC bisects BD, prove that BD bisects AC. 2. **Understand the given information:** - AB \parallel CD means lines AB and CD are parallel. - AC bisects BD means point E (intersection of AC and BD) divides BD into two equal segments, so BE = ED. 3. **Goal:** Prove that BD bisects AC, i.e., E also divides AC into two equal segments, so AE = EC. 4. **Use properties of parallel lines and transversals:** Since AB \parallel CD and AC intersects them, alternate interior angles are equal: $$\angle BAE = \angle DCE$$ 5. **Triangles ABE and CDE:** - BE = ED (given, since AC bisects BD) - AB \parallel CD (given) - \angle BAE = \angle DCE (alternate interior angles) 6. **By ASA (Angle-Side-Angle) congruence criterion, triangles ABE and CDE are congruent:** $$\triangle ABE \cong \triangle CDE$$ 7. **From congruence, corresponding parts are equal:** $$AE = CE$$ 8. **Conclusion:** Since E divides AC into two equal parts, BD bisects AC. **Final answer:** BD bisects AC.