In ΔABC shown below, segment DE is parallel to segment AC:Triangles ABC and DBE where DE is parallel to ACThe 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 Reason1. Line segment DE is parallel to line segment AC 1. Given2. 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 Equality5. 5.6. BD over BA equals BE over BC 6. Converse of the Side-Side-Side Similarity TheoremWhich statement and reason accurately completes the proof? 3. ∠BDE ≅ ∠BAC; Corresponding Angles Postulate5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate 3. ΔBDE ~ ΔBAC; Corresponding Angles Postulate5. ∠BDE ~ ∠BAC; Angle-Angle (AA) Similarity Postulate 3. ∠BDE ≅ ∠BAC; Congruent Angles Postulate5. ΔBDE ~ ΔBAC; Angle-Angle (AA) Similarity Postulate 3. ∠BDE ≅ ∠BAC; Congruent Angles Postulate5. ΔBDE ~ ΔBAC; Side-Angle-Side (SAS) Similarity Postulate
Accepted Solution
A:
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 BAC1. 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