In ΔABC shown below, segment DE is parallel to segment AC: Triangles ABC and DBE where DE is parallel to AC

The following two-column proof with missing statements and reasons proves that if a line parallel to one side of a triangle also intersects the other two sides, the line divides the sides proportionally:

StatementReason
1. Line segment DE is parallel to line segment AC1. Given
2. Line segment AB is a transversal that intersects two parallel lines.2. Conclusion from Statement 1.
3.3.
4. ∠B ≅ ∠B4. Reflexive Property of Equality
5.5.
6. BD over BA equals BE over BC6. Converse of the Side-Side-Side Similarity Theorem

Which statement and reason accurately completes the proof?
3. ∠BDE ≅ ∠BAC; Corresponding Angles Postulate
5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate

3. ΔBDE ~ ΔBAC; Corresponding Angles Postulate
5. ∠BDE ~ ∠BAC; Angle-Angle (AA) Similarity Postulate

3. ∠BDE ≅ ∠BAC; Congruent Angles Postulate
5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate

3. ∠BDE ≅ ∠BAC; Congruent Angles Postulate
5. ΔBDE ~ ΔBAC; Side-Angle-Side (SAS) Similarity Postulate

Also sorry, I couldn't get the picture of the triangle to paste, I don't know if you truly need that to answer the question but if you do, then its okay if you can't help me.