Elementary Geometry for College Students (7th Edition) Clone

- From the given information, we can deduce $\angle 1\cong\angle 2$ and $\angle 3\cong\angle 4$ - Then, prove that $\triangle MQP\cong\triangle PNM$ by method ASA - Then, by CPCTC, $\overline{MQ}\cong\overline{PN}$
*PLANNING: - From the given information, we can deduce $\angle 1\cong\angle 2$ and $\angle 3\cong\angle 4$ - Then, prove that $\triangle MQP\cong\triangle PNM$ - Then, by CPCTC, $\overline{MQ}\cong\overline{PN}$ 1) $\overline{MN}\parallel\overline{QP}$ and $\overline{MQ}\parallel\overline{NP}$ (Given) 2) $\angle 1\cong\angle 2$ and $\angle 3\cong\angle 4$ (if 2 lines are parallel, then the alternate interior angles for these 2 lines are congruent) 3) $\overline{MP}\cong\overline{PM}$ (Identity) So now we have 2 angles and the included side of $\triangle MQP$ are congruent with 2 corresponding angles and the included side of $\triangle PNM$ 4) $\triangle MQP\cong\triangle PNM$ (ASA) 5) $\overline{MQ}\cong\overline{PN}$ (CPCTC) 