Question

given: n || m and <1 = <2 prove: p || r n 1 m 3 2 r p n is parallel to m, <1 is congruent to <2 <1 is congruent to <3 <2 is congruent to <3 p is parallel to r given symmetric property converse of corresponding angles converse of alternate interior angles alternate exterior angles are congruent converse of alternate interior angles transitive/substitution property alternate interior angles are congruent corresponding angles are congruent reflexive property

Explanation:

Step1: Given information

We are given that $n\parallel m$ and $\angle1=\angle2$.

Step2: Use corresponding - angles property

Since $n\parallel m$, by the corresponding - angles property, $\angle1$ and $\angle3$ are corresponding angles, so $\angle1\cong\angle3$.

Step3: Use transitive property

We know $\angle1 = \angle2$ (given) and $\angle1=\angle3$ (from corresponding - angles of parallel lines). By the transitive/substitution property, $\angle2\cong\angle3$.

Step4: Use converse of alternate interior angles

If $\angle2\cong\angle3$, then by the converse of alternate interior angles, $p\parallel r$.

Answer:

$n$ is parallel to $m$, $\angle1$ is congruent to $\angle3$ because corresponding angles are congruent when lines are parallel. Since $\angle1=\angle2$ (given), by the transitive/substitution property $\angle2\cong\angle3$. Then, by the converse of alternate interior angles, $p\parallel r$.