select the appropriate rephrased statement for weis proof.
choose 1 answer:
a in △pqr, if $\overline{st} \parallel \overline{qr}$, then $ps = sq$ and $pt = tr$.
b in △pqr, if $ps = sq$ and $pt = tr$, then $\overline{st} \parallel \overline{qr}$.
c in △pqr, if $\overline{st} \parallel \overline{qr}$, then $ps = pt$ and $pq = pr$.
d in △pqr, if $ps = pt$ and $pq = pr$, then $\overline{st} \parallel \overline{qr}$.

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# Brief Explanations:
To solve this, we recall the Midline Theorem (also known as the Midsegment Theorem) in triangles. The theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side. So, if $ S $ is the midpoint of $ PQ $ (i.e., $ PS = SQ $) and $ T $ is the midpoint of $ PR $ (i.e., $ PT = TR $), then $ ST \parallel QR $.

- Option A: Assumes that parallelism implies midpoints, but the Midline Theorem is the converse (midpoints imply parallelism), so A is incorrect.
- Option B: Matches the Midline Theorem (midpoints of two sides imply the segment is parallel to the third side), so B is correct.
- Option C: Incorrectly relates parallelism to equal segments $ PS = PT $ and $ PQ = PR $, which is not a valid geometric relationship here.
- Option D: Incorrectly uses $ PS = PT $ and $ PQ = PR $ to imply parallelism, which does not follow from the Midline Theorem or other standard triangle theorems.

# Answer:
B. In $ \triangle PQR $, if $ PS = SQ $ and $ PT = TR $, then $ \overline{ST} \parallel \overline{QR} $

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