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:

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. 5.
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 ≅ ∠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

Question

contestada

Respuesta :

inchu420 inchu420 · Answer 1 · 2019-12-12T06:51:07+01:00

Answer:

Proof withe statement is Given below.

Step-by-step explanation:

Given:

Triangles ABC and DBE where DE is parallel to AC

To Prove:

[tex]\frac{BD}{BA} =\frac{BE}{BC}[/tex]

Proof:

In Δ DBE and Δ BAC

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. ∠B ≅ ∠B 4. Reflexive Property of Equality

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