Question

given: lines p and q are parallel and r is a transversal. prove: ∠2 ≅ ∠7
diagram: lines p (horizontal, left-right) and q (horizontal, left-right) are parallel, transversal r intersects them, forming angles 1,2,3,4 (with p) and 5,6,7,8 (with q).
statements | reasons
--- | ---

1. ( p parallel q ), ( r ) is a transv. | 1. given
2. a | 2. vert. ( angle )’s ( cong )
3. b | 3. corr. ( angle )’s thm.
4. ( angle 2 cong angle 7 ) | 4. transitive prop.

which statements could complete the proof?
a: checkbox
b: checkbox

Explanation:

Step1: Analyze Statement A (Vertical Angles)

Vertical angles are congruent. For line \( p \) and transversal \( r \), \( \angle 2 \) and \( \angle 3 \)? Wait, no—wait, \( \angle 2 \) and \( \angle 4 \)? No, wait, looking at the diagram, \( \angle 2 \) and \( \angle 3 \) are adjacent, but \( \angle 1 \) and \( \angle 3 \) are vertical? Wait, no, let's re-examine. Wait, the vertical angle for \( \angle 2 \): actually, \( \angle 2 \) and \( \angle 3 \)? No, \( \angle 1 \) and \( \angle 3 \) are vertical, \( \angle 2 \) and \( \angle 4 \) are vertical? Wait, no, maybe I messed up. Wait, the reason for A is "vert. \( \angle \)'s \( \cong \)", so we need a pair of vertical angles. Wait, maybe \( \angle 2 \cong \angle 3 \)? No, wait, no—wait, the goal is to connect \( \angle 2 \) to \( \angle 7 \). Let's think: \( \angle 2 \) and \( \angle 6 \) are corresponding angles (since \( p \parallel q \), transversal \( r \)), but no, the reason for B is "corr. \( \angle \)'s thm." So corr. angles would be, for example, \( \angle 2 \cong \angle 6 \) (since \( p \parallel q \), transversal \( r \), \( \angle 2 \) and \( \angle 6 \) are corresponding). But then, what about vertical angles? Wait, \( \angle 6 \) and \( \angle 7 \) are vertical angles? Wait, no, \( \angle 6 \) and \( \angle 8 \) are vertical? Wait, the diagram: \( \angle 5, \angle 6, \angle 7, \angle 8 \) around the intersection with \( q \). So \( \angle 6 \) and \( \angle 8 \) are vertical, \( \angle 5 \) and \( \angle 7 \) are vertical. Wait, maybe I got the vertical angles wrong. Let's start over.

Given \( p \parallel q \), transversal \( r \). We need to prove \( \angle 2 \cong \angle 7 \).

Step 2: For Statement A (Reason: vert. \( \angle \)'s \( \cong \)): We need a pair of vertical angles. Let's see, \( \angle 2 \) and \( \angle 3 \)? No, \( \angle 1 \) and \( \angle 3 \) are vertical, \( \angle 2 \) and \( \angle 4 \) are vertical? Wait, no, maybe \( \angle 2 \cong \angle 3 \)? No, that's not right. Wait, maybe \( \angle 2 \cong \angle 4 \)? No, vertical angles are opposite each other. Wait, the intersection of \( r \) and \( p \): angles \( 1, 2, 3, 4 \). So \( \angle 1 \) and \( \angle 3 \) are vertical, \( \angle 2 \) and \( \angle 4 \) are vertical. At the intersection of \( r \) and \( q \): angles \( 5, 6, 7, 8 \). \( \angle 5 \) and \( \angle 7 \) are vertical, \( \angle 6 \) and \( \angle 8 \) are vertical.

Now, the reason for B is "corr. \( \angle \)'s thm." (corresponding angles theorem), which states that if two parallel lines are cut by a transversal, corresponding angles are congruent. So \( \angle 2 \) and \( \angle 6 \) are corresponding angles (since \( p \parallel q \), transversal \( r \), same position relative to \( p, q \) and transversal \( r \)). So \( \angle 2 \cong \angle 6 \) (corr. angles thm.). Then, for Statement A, the reason is vertical angles congruent, so we need \( \angle 6 \cong \angle 7 \) (since \( \angle 6 \) and \( \angle 7 \) are vertical angles). Then, by transitive property, \( \angle 2 \cong \angle 6 \) and \( \angle 6 \cong \angle 7 \) implies \( \angle 2 \cong \angle 7 \).

So:

- Statement A (Reason: vert. \( \angle \)'s \( \cong \)): \( \angle 6 \cong \angle 7 \) (because \( \angle 6 \) and \( \angle 7 \) are vertical angles)
- Statement B (Reason: corr. \( \angle \)'s thm.): \( \angle 2 \cong \angle 6 \) (because \( p \parallel q \), transversal \( r \), so corresponding angles \( \angle 2 \) and \( \angle 6 \) are congruent)

Let's verify:

1. \( p \parallel q \), \( r \) is transv. (given)
2. \( \angle 6 \cong \angle 7 \) (vert. \( \angle \)'s \( \cong \)) → Statement A
3. \( \angle 2 \cong \angle 6 \) (corr. \( \angle \)'s thm.) → Statement B
4. \( \angle 2 \cong \angle 7 \) (transitive prop: if \( \angle 2 \cong \angle 6 \) and \( \angle 6 \cong \angle 7 \), then \( \angle 2 \cong \angle 7 \))

Yes, that works. So:

A: \( \angle 6 \cong \angle 7 \)

B: \( \angle 2 \cong \angle 6 \)

Answer:

A: \( \boldsymbol{\angle 6 \cong \angle 7} \)
B: \( \boldsymbol{\angle 2 \cong \angle 6} \)