Answer :
To complete the two-column proof, we need to add the statement and reason in step 5. Here is the detailed solution.
Given:
1. \( \overline{DE} \parallel \overline{AC} \) (Given)
Based on this, we know:
2. \( \overline{AB} \) is a transversal that intersects two parallel lines. (Conclusion from Statement 1).
From the Corresponding Angles Postulate, we have:
3. \( \angle BDE \cong \angle BAC \) (Corresponding Angles Postulate)
By the Reflexive Property of Equality, we also have:
4. \( \angle B \cong \angle B \) (Reflexive Property of Equality)
Now, since we have two pairs of corresponding angles that are congruent (\( \angle BDE \cong \angle BAC \) and \( \angle B \cong \angle B \)), by the Angle-Angle (AA) Similarity Postulate, we can say:
5. \( \triangle BDE \sim \triangle BAC \) (Angle-Angle (AA) Similarity Postulate)
This ensures that the corresponding sides of these triangles are proportional.
Lastly, based on the similarity of the triangles, we can conclude:
6. \( \frac{BD}{BA} = \frac{BE}{BC} \) (Converse of the Side-Side-Side Similarity Theorem)
So, the completed two-column proof looks like this:
\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{ Statement } & \multicolumn{1}{|c|}{ Reason } \\
\hline 1. [tex]$\overline{ DE } \| \overline { AC }$[/tex] & 1. Given \\
\hline 2. [tex]$\overline{ AB }$[/tex] is a transversal that intersects two parallel lines. & 2. Conclusion from Statement 1. \\
\hline 3. [tex]$\angle BDE \cong \angle BAC$[/tex] & 3. Corresponding Angles Postulate \\
\hline 4. [tex]$\angle B \cong \angle B$[/tex] & 4. Reflexive Property of Equality \\
\hline 5. [tex]$\triangle B D E \sim \triangle B A C$[/tex] & 5. Angle-Angle (AA) Similarity Postulate \\
\hline 6. [tex]$\frac{ BD }{ BA } = \frac{ BE }{ BC }$[/tex] & 6. Converse of the Side-Side-Side Similarity Theorem \\
\hline
\end{tabular}
Therefore, the accurate completion for statement 5 is:
[tex]\[ \boxed{5. \triangle BDE \sim \triangle BAC; \text{Angle-Angle (AA) Similarity Postulate}} \][/tex]
Given:
1. \( \overline{DE} \parallel \overline{AC} \) (Given)
Based on this, we know:
2. \( \overline{AB} \) is a transversal that intersects two parallel lines. (Conclusion from Statement 1).
From the Corresponding Angles Postulate, we have:
3. \( \angle BDE \cong \angle BAC \) (Corresponding Angles Postulate)
By the Reflexive Property of Equality, we also have:
4. \( \angle B \cong \angle B \) (Reflexive Property of Equality)
Now, since we have two pairs of corresponding angles that are congruent (\( \angle BDE \cong \angle BAC \) and \( \angle B \cong \angle B \)), by the Angle-Angle (AA) Similarity Postulate, we can say:
5. \( \triangle BDE \sim \triangle BAC \) (Angle-Angle (AA) Similarity Postulate)
This ensures that the corresponding sides of these triangles are proportional.
Lastly, based on the similarity of the triangles, we can conclude:
6. \( \frac{BD}{BA} = \frac{BE}{BC} \) (Converse of the Side-Side-Side Similarity Theorem)
So, the completed two-column proof looks like this:
\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{ Statement } & \multicolumn{1}{|c|}{ Reason } \\
\hline 1. [tex]$\overline{ DE } \| \overline { AC }$[/tex] & 1. Given \\
\hline 2. [tex]$\overline{ AB }$[/tex] is a transversal that intersects two parallel lines. & 2. Conclusion from Statement 1. \\
\hline 3. [tex]$\angle BDE \cong \angle BAC$[/tex] & 3. Corresponding Angles Postulate \\
\hline 4. [tex]$\angle B \cong \angle B$[/tex] & 4. Reflexive Property of Equality \\
\hline 5. [tex]$\triangle B D E \sim \triangle B A C$[/tex] & 5. Angle-Angle (AA) Similarity Postulate \\
\hline 6. [tex]$\frac{ BD }{ BA } = \frac{ BE }{ BC }$[/tex] & 6. Converse of the Side-Side-Side Similarity Theorem \\
\hline
\end{tabular}
Therefore, the accurate completion for statement 5 is:
[tex]\[ \boxed{5. \triangle BDE \sim \triangle BAC; \text{Angle-Angle (AA) Similarity Postulate}} \][/tex]
