In ΔABC shown below, BD/BA = BE/BC:
Triangle ABC with segment DE intersecting sides AB and BC respectively.
The following flowchart proof with missing statements and reasons proves that if a line intersects two sides of a triangle and divides these sides proportionally, the line is parallel to the third side:
Top path, by Given: BD/BA = BE/BC. By Side-Angle-Side Similarity Postulate, triangle ABC is similar to triangle DBE. By space labeled by 2, space labeled by 1 occurs. By Converse of the Corresponding Angles Postulate, line segment DE is parallel to line segment AC. Bottom path, by Reflexive Property of Equality, angle B is congruent to angle B. By Side-Angle-Side Similarity Postulate, triangle ABC is similar to triangle DBE. By space labeled by 2, space labeled by 1 occurs. By Converse of the Corresponding Angles Postulate, line segment DE is parallel to line segment AC.
Which reason can be used to fill in the numbered blank space?
1. ∠BDE ≅ ∠BAC
2. Corresponding Angles Postulate
1. ∠BDE ≅ ∠BAC
2. Corresponding Parts of Similar Triangles
1. ∠BDE ≅ ∠BCA
2. Alternate Exterior Theorem
1. ∠BDE ≅ ∠BCA
2. Corresponding Parts of Similar Triangles