Kayla is attempting to prove MQ=12AP . She has already shown that MN=12AC and MN¯¯¯¯¯¯¯¯¯¯ ∥ AC¯¯¯¯¯¯¯¯ by applying the Triangle Midsegment Theorem. Kayla now wants to prove that △MBQ∼△ABP . She notices that △MBQ and △ABP share the angle ∠MBQ . If Kayla can prove ∠BQM≅∠BPA , she can conclude that △MBQ∼△ABP by the AAA Similarity Theorem. Which of the following is the correct reasoning to prove ∠BQM≅∠BPA ?

Question

Because line MN is parallel to line AC and alternate exterior angles of parallel lines are congruent, ∠BQM≅∠BPA.

Because line MN is parallel to line AC and corresponding angles of parallel lines are congruent, ∠BQM≅∠BPA

Because triangle MBQ is similar triangle ABP and corresponding angles of similar triangles are congruent, angle MBQ≅ angle BPA

Because triangle MBN is similar to Triangle ABC and Q is midpoint of line MN , ∠BQM≅∠BPA

Answer 1

To prove that \(\angle BQM \cong \angle BPA\) based on the relationship between lines \(MN\) and \(AC\), the correct reasoning would be:

Because line MN is parallel to line AC and alternate exterior angles of parallel lines are congruent, \(\angle BQM \cong \angle BPA\).

This conclusion uses the fact that if two lines are parallel, then the alternate exterior angles formed by a transversal cutting through them are congruent. In this case, \(MN\) and \(AC\) are parallel, and rays \(BQ\) and \(AP\) can be considered as transversals. Therefore, the angles \(\angle BQM\) and \(\angle BPA\) are indeed congruent.