Question

b. can it also be proved that $overline{pq}congoverline{rs}$ if $overline{pr}congoverline{qs}$? explain. yes, the segment addition postulate can be used to show that $pr = pq + qr$ and $qs=qr + rs$. both equations can be solved for qr, and substituting select choice for select choice will lead to $overline{pq}congoverline{rs}$.

Explanation:

Step1: Apply segment - addition postulate

$PR = PQ+QR$ and $QS = QR + RS$

Step2: Solve for $QR$

From $PR = PQ+QR$, we get $QR=PR - PQ$. From $QS = QR + RS$, we get $QR=QS - RS$.

Step3: Substitute

Since $PR\cong QS$, then $PR = QS$. Substituting $QR$ expressions: $PR - PQ=QS - RS$. Because $PR = QS$, we can subtract them from both sides of the equation, resulting in $PQ = RS$, so $\overline{PQ}\cong\overline{RS}$.

Answer:

Yes, the Segment Addition Postulate can be used to show that $PR = PQ+QR$ and $QS = QR + RS$. Both equations can be solved for $QR$, and substituting $PR - PQ$ for $QS - RS$ (since $PR = QS$) will lead to $\overline{PQ}\cong\overline{RS}$.