Question

a student is checking whether the design she draw is symmetric. can she determine whether mn and pn are the same length? explain.
choose the correct answer below
a. no. in order to show \\( \overline { m n } \cong \overline { p n } , \triangle s m n \\) must be congruent to \\( \triangle u n p \\). the only criterion that can be found for \\( \triangle s m n \\) and \\( \triangle u n p \\) is ssa, which is not a congruence criterion.
b. yes. by sas, \\( \triangle r s t \cong \triangle q u t \\), and therefore \\( \overline { s t } \cong \overline { u t } \\). so \\( \overline { r u } \cong \overline { q s } \\), and by asa, \\( \triangle r u n \cong \triangle q s n \\). therefore, \\( \overline { s n } \cong \overline { u n } \\) and \\( \triangle s m n \cong \triangle u n p \\) by ssa, which means that \\( \overline { m n } \cong \overline { p n } \\).
c. no. in order to show \\( \overline { m n } \cong \overline { p n } , \triangle r n p \\) must be shown to be congruent to \\( \triangle q n m \\), and that cannot be done by any congruence criterion.
d. yes. by sas, \\( \triangle r s t \cong \triangle q u t \\), and therefore \\( \overline { s t } \cong \overline { u t } \\). so \\( \overline { r p } \cong \overline { q m } \\), and by asa, \\( \triangle r n p \cong \triangle q n m \\), which means that \\( \overline { m n } \cong \overline { p n } \\).

Explanation:

Step1: Analyze option A

SSA (Side - Side - Angle) is not a valid congruence criterion. So, just because we want to show \( \overline{MN}\cong\overline{PN}\) and we have some triangles (\(\triangle SMN\) and \(\triangle UNP\)) with SSA - like information does not mean we can use it to prove congruence. So, we cannot use SSA to show \( \overline{MN}\cong\overline{PN}\) through \(\triangle SMN\) and \(\triangle UNP\).

Step2: Analyze option B

If \(\triangle RST\cong\triangle QUT\) by SAS (Side - Angle - Side: assume \(RS = QU\), \(\angle RST=\angle QUT\), \(ST = UT\)), then \(RU = QS\) (by corresponding parts of congruent triangles). If we consider \(\triangle RUN\) and \(\triangle QSN\), and assume we can use ASA (Angle - Side - Angle) to show \(\triangle RUN\cong\triangle QSN\) (say \(\angle RUN=\angle QSN\), \(RU = QS\), \(\angle RNU=\angle QNU\)), then \(SN = UN\). But SSA (\(\triangle SMN\) and \(\triangle UNP\): \(SN = UN\), \(\angle SMN=\angle UNP\), \(MN\) and \(PN\) are the sides we want to check) is not a valid congruence criterion. So, we cannot say \(MN\cong PN\) from \(\triangle RST\cong\triangle QUT\) and then using some wrong congruence steps.

Step3: Analyze option C

To show \( \overline{MN}\cong\overline{PN}\), we need to have congruent triangles. If we consider \(\triangle RNP\) and \(\triangle QNM\), there is no information given about the angles and sides that would satisfy a valid congruence criterion (like SSS - Side - Side - Side, SAS, ASA - Angle - Side - Angle, AAS - Angle - Angle - Side). Just wanting to show \( \overline{MN}\cong\overline{PN}\) for \(\triangle RNP\) and \(\triangle QNM\) without a valid congruence criterion (not SSA, not SSS, not SAS, not ASA, not AAS) is not possible.

Step4: Analyze option D

If \(\triangle RST\cong\triangle QUT\) by SAS (\(RS = QU\), \(\angle RST=\angle QUT\), \(ST = UT\)), then \(RT = QT\). Also, if we assume \(RP = QM\) (by some other corresponding parts or given information) and \(\angle RPT=\angle QMT\) (vertical angles or from the symmetry of the figure). Then, in \(\triangle RPN\) and \(\triangle QMN\), we have \(RP = QM\), \(\angle RPN=\angle QMN\), \(PN = MN\) (by ASA: if we can show two angles and the included side). Wait, actually, if \(\triangle RST\cong\triangle QUT\) (SAS: \(RS = QU\), \(\angle RST=\angle QUT\), \(ST = UT\)), then \(RU = QS\) (corresponding parts). If we consider \(\triangle RUN\) and \(\triangle QSN\) (ASA: assume \(\angle RUN=\angle QSN\), \(RU = QS\), \(\angle RNU=\angle QNU\)), then \(SN = UN\). Now, in \(\triangle SMN\) and \(\triangle UNP\), if we assume \(\angle SMN=\angle UNP\) (from symmetry) and \(SN = UN\), and if we can show \(MN = PN\) using SAS (if \(\angle MNS=\angle PNU\), \(SN = UN\), \(MN\) and \(PN\) with the included angles). But actually, if \(\triangle RST\cong\triangle QUT\) (SAS), then \(ST = UT\). If we consider \(\triangle RNP\) and \(\triangle QNM\) (assuming \(RP = QM\) from \(RT - PT=QT - MT\) and \(\angle RPN=\angle QMN\) (from the overall symmetry of the figure) and \(\angle RNP=\angle QNM\) (vertical angles), then \(\triangle RNP\cong\triangle QNM\) (ASA) and \(MN = PN\).

Answer:

D. Yes By SAS, \(\triangle RST\cong\triangle QUT\), and therefore \(ST\cong UT\). So \(RP\cong QM\), and by ASA, \(\triangle RNP\cong\triangle QNM\), which means that \(MN\cong PN\)