Question

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 proves that if a line parallel to one side of a triangle also intersects the other two sides, the line divides the sides proportionally.


Statement Reason
1. Line segment DE is parallel to line segment AC 1. Given
2. Line segment AB is a transversal that intersects two parallel lines. 2. Conclusion from Statement 1.
3. 3.
4. ∠B ≅ ∠B 4. Reflexive Property of Equality
5. ΔABC ~ ΔDBE 5. Angle-Angle (AA) Similarity Postulate
6. BD over BA equals BE over BC 6. Converse of the Side-Side-Side Similarity Theorem


Which statement and reason accurately completes the proof?
3. ∠BDE ≅ ∠ABC; Corresponding Angles Postulate
3. ∠BDE ≅ ∠ABC; Alternate Interior Angles Theorem
3. ∠BDE ≅ ∠BAC; Corresponding Angles Postulate
3. ∠BDE ≅ ∠BAC; Alternate Interior Angles Theorem

Answer 1

The correct statement to complete the proof is that ∠BDE is congruent to ∠BAC, and the reason is the Corresponding Angles Postulate.

The statement and reason that accurately completes the proof are:

∠BDE ≅ ∠BAC; Corresponding Angles Postulate.

This is because when two lines are parallel, like DE and AC, and they are cut by a transversal, in this case, AB, the corresponding angles are congruent.

Following this postulate, if DE is parallel to AC and AB is a transversal that intersects DE at D and AC at B, then angle BDE must be congruent to angle BAC.