Answer :
Alright! Let's break down the steps of the proof in a detailed and sequential manner to complete the two-column proof provided:
1. Statement: [tex]\( \overline{DE} \parallel \overline{AC} \)[/tex]
Reason: Given
2. Statement: [tex]\( \overline{AB} \)[/tex] is a transversal that intersects two parallel lines.
Reason: Conclusion from Statement 1
3. Statement: [tex]\( \angle BDE \cong \angle BAC \)[/tex]
Reason: Corresponding Angles Postulate. When a transversal intersects two parallel lines, the corresponding angles are congruent.
4. Statement: [tex]\( \angle B \cong \angle B \)[/tex]
Reason: Reflexive Property of Equality. Any angle is congruent to itself.
5. Statement: [tex]\( \triangle BDE \sim \triangle BAC \)[/tex]
Reason: Angle-Angle (AA) Similarity Postulate. Two triangles are similar if two pairs of corresponding angles are congruent.
6. Statement: [tex]\( \frac{BD}{BA} = \frac{BE}{BC} \)[/tex]
Reason: Corresponding sides of similar triangles are in proportion (Converse of the Side-Side-Side Similarity Theorem).
To address the specific options given for step 5 and complete the proof accurately:
- Option 1: 5. [tex]\( \triangle BDE \sim \triangle BAC \)[/tex]; Angle-Angle (AA) Similarity Postulate
- Option 2: 5. [tex]\( \angle BDE \sim \angle BAC \)[/tex]; Angle-Angle (AA) Similarity Postulate
- Option 3: 5. [tex]\( \angle BDE \sim \angle BAC \)[/tex]; Side-Angle-Side (SAS) Similarity Postulate
- Option 4: 5. [tex]\( \triangle BDE \sim \triangle BAC \)[/tex]; Side-Angle-Side (SAS) Similarity Postulate
Given the reasoning and the nature of the proof, the correct statement and reason that accurately complete the proof are:
### 5. [tex]\( \triangle BDE \sim \triangle BAC \)[/tex]; Angle-Angle (AA) Similarity Postulate
This is because we are using the fact that two pairs of corresponding angles are congruent to establish that the triangles are similar, rather than using the Side-Angle-Side postulate.
So, the completed proof looks like this:
\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{Statement} & \multicolumn{1}{|c|}{Reason} \\
\hline 1. [tex]\( \overline{DE} \parallel \overline{AC} \)[/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]\( \angle B \)[/tex] congruent [tex]\( \angle B \)[/tex] & 4. Reflexive Property of Equality \\
\hline 5. [tex]\( \triangle BDE \sim \triangle BAC \)[/tex] & 5. Angle-Angle (AA) Similarity Postulate \\
\hline 6. [tex]\( \frac{BD}{BA} = \frac{BE}{BC} \)[/tex] & 6. Corresponding sides of similar triangles are in proportion (Converse of the Side-Side-Side Similarity Theorem) \\
\hline
\end{tabular}
This proof demonstrates that if a line parallel to one side of a triangle also intersects the other two sides, it will divide the sides proportionally.
1. Statement: [tex]\( \overline{DE} \parallel \overline{AC} \)[/tex]
Reason: Given
2. Statement: [tex]\( \overline{AB} \)[/tex] is a transversal that intersects two parallel lines.
Reason: Conclusion from Statement 1
3. Statement: [tex]\( \angle BDE \cong \angle BAC \)[/tex]
Reason: Corresponding Angles Postulate. When a transversal intersects two parallel lines, the corresponding angles are congruent.
4. Statement: [tex]\( \angle B \cong \angle B \)[/tex]
Reason: Reflexive Property of Equality. Any angle is congruent to itself.
5. Statement: [tex]\( \triangle BDE \sim \triangle BAC \)[/tex]
Reason: Angle-Angle (AA) Similarity Postulate. Two triangles are similar if two pairs of corresponding angles are congruent.
6. Statement: [tex]\( \frac{BD}{BA} = \frac{BE}{BC} \)[/tex]
Reason: Corresponding sides of similar triangles are in proportion (Converse of the Side-Side-Side Similarity Theorem).
To address the specific options given for step 5 and complete the proof accurately:
- Option 1: 5. [tex]\( \triangle BDE \sim \triangle BAC \)[/tex]; Angle-Angle (AA) Similarity Postulate
- Option 2: 5. [tex]\( \angle BDE \sim \angle BAC \)[/tex]; Angle-Angle (AA) Similarity Postulate
- Option 3: 5. [tex]\( \angle BDE \sim \angle BAC \)[/tex]; Side-Angle-Side (SAS) Similarity Postulate
- Option 4: 5. [tex]\( \triangle BDE \sim \triangle BAC \)[/tex]; Side-Angle-Side (SAS) Similarity Postulate
Given the reasoning and the nature of the proof, the correct statement and reason that accurately complete the proof are:
### 5. [tex]\( \triangle BDE \sim \triangle BAC \)[/tex]; Angle-Angle (AA) Similarity Postulate
This is because we are using the fact that two pairs of corresponding angles are congruent to establish that the triangles are similar, rather than using the Side-Angle-Side postulate.
So, the completed proof looks like this:
\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{Statement} & \multicolumn{1}{|c|}{Reason} \\
\hline 1. [tex]\( \overline{DE} \parallel \overline{AC} \)[/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]\( \angle B \)[/tex] congruent [tex]\( \angle B \)[/tex] & 4. Reflexive Property of Equality \\
\hline 5. [tex]\( \triangle BDE \sim \triangle BAC \)[/tex] & 5. Angle-Angle (AA) Similarity Postulate \\
\hline 6. [tex]\( \frac{BD}{BA} = \frac{BE}{BC} \)[/tex] & 6. Corresponding sides of similar triangles are in proportion (Converse of the Side-Side-Side Similarity Theorem) \\
\hline
\end{tabular}
This proof demonstrates that if a line parallel to one side of a triangle also intersects the other two sides, it will divide the sides proportionally.
