figure abcd is a parallelogram. which sequence could be used to prove that ad = bc? first prove △abc is congruent to △cda, and then state $ \overline { ad } $ and $ \overline { bc } $ are corresponding sides of the triangles. first prove △abc is similar to △cda, and then state $ \overline { ad } $ and $ \overline { bc } $ are opposite sides of the parallelograms. first prove □abcd is congruent to □cdab, and then state $ \overline { ad } $ and $ \overline { bc } $ are corresponding sides of two parallelograms. first prove □abcd is similar to □cdab, and then state $ \overline { ad } $ and $ \overline { bc } $ are opposite sides of the parallelograms.

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# Brief Explanations:
- Congruent triangles have corresponding sides equal. If we can prove that $\triangle ABC\cong\triangle CDA$, then by the property of congruent triangles (corresponding parts of congruent triangles are congruent - CPCTC), we can say that $AD = BC$ as they are corresponding sides of the congruent triangles $\triangle ABC$ and $\triangle CDA$.
- Similar triangles have proportional sides, not necessarily equal. So proving similarity ($\triangle ABC\sim\triangle CDA$)  won't directly give $AD = BC$.
- A parallelogram is a single figure, and we don't prove a parallelogram $ABCD$ congruent or similar to itself ($CDAB$ is the same parallelogram $ABCD$ just named in a different vertex - order). The key is to use triangle congruence within the parallelogram.

# Answer:
First prove $\triangle ABC$ is congruent to $\triangle CDA$, and then state $\overline{AD}$ and $\overline{BC}$ are corresponding sides of the triangles.

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