Question

use the properties of rigid motions to complete the proof. given: $overline{pq}congoverline{qr}$ and $mangle pqscong mangle rqs$. prove: $overline{qs}perpoverline{pr}$ and $overline{ps}congoverline{sr}$. reflections preserve angle measures and distances, so the image of $overline{qp}$ lies along $overline{qr}$ and has the same length as $overline{qr}$.

Explanation:

Step1: Recall reflection property

A reflection is a rigid - motion. Rigid motions preserve distance. Given $\overline{PQ}\cong\overline{QR}$, when we reflect $\triangle QPS$ across $\overline{QS}$, since reflections preserve distances, the image of $\overline{QP}$ (which is $\overline{QR}$) has the same length as $\overline{QR}$. Also, given $m\angle PQS\cong m\angle RQS$, the line of reflection $\overline{QS}$ is the perpendicular bisector of $\overline{PR}$ (by the property of angle - bisectors and congruent triangles). And since $\overline{PS}\cong\overline{SR}$ (given), and $\overline{QS}\perp\overline{PR}$ (to be proved). In $\triangle PQS$ and $\triangle RQS$, we have $\overline{PQ}\cong\overline{QR}$, $m\angle PQS\cong m\angle RQS$, and $\overline{QS}=\overline{QS}$ (common side). By the Side - Angle - Side (SAS) congruence criterion, $\triangle PQS\cong\triangle RQS$. Then, corresponding parts of congruent triangles are congruent. Since $\angle QSP$ and $\angle QSR$ are corresponding angles and $\angle QSP+\angle QSR = 180^{\circ}$ (linear pair), and $\triangle PQS\cong\triangle RQS$, $\angle QSP=\angle QSR = 90^{\circ}$, so $\overline{QS}\perp\overline{PR}$.

Answer:

The proof is completed by using the properties of rigid motions (reflection preserves distance and angle measure), the Side - Angle - Side (SAS) congruence criterion, and the properties of corresponding parts of congruent triangles.