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. ∠bde ≅ ∠bac 3. corresponding angles postulate 4. 4. 5. Δbde Δbac 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? 4. ∠a ≅ ∠a; reflexive property of equality 4. ∠b ≅ ∠b; reflexive property of equality 4. ∠b ≅ ∠c; isosceles triangle theorem 4. ∠a ≅ ∠c; isosceles triangle theorem

1) ∠a ≅ ∠a; reflexive property of equality
2) ∠b ≅ ∠b; reflexive property of equality
3) ∠b ≅ ∠c; isosceles triangle theorem
4) ∠a ≅ ∠c; isosceles triangle theorem

Answer 1

The correct completion for the proof that a parallel line divides the sides of a triangle proportionally involves the reflexive property of equality, indicating that angle A is congruent to itself.

Step-by-step explanation:

The question is about completing a two-column proof that demonstrates the property of a line dividing the sides of a triangle proportionally when it is parallel to one of the triangle's sides. Specifically, it concerns the missing reason for statement 4 in proving that triangle ΔBDE is similar to triangle ΔBAC. The correct statement and reason to complete the proof are '∠A ≅ ∠A; reflexive property of equality' because each angle in a triangle is congruent to itself due to the reflexive property of equality which is necessary to establish the Angle-Angle (AA) similarity between the triangles.