- Identify what congruent alternate interior angles imply about the two lines cut by the transversal. - Use the converse theorem that turns an angle relationship into parallelism.

1. Congruent alternate interior angles satisfy the converse of the alternate-interior-angles theorem. 2. Therefore, the two lines cut by the transversal are parallel. 3. Since \(g\) passes through \(P\), the construction has produced the required parallel through \(P\).

It verifies that \(g\parallel\ell\).

- A single perpendicular changes the direction by \(90^\circ\). - Two coplanar lines perpendicular to the same line are parallel.

1. Choice A produces a perpendicular, not a parallel. 2. In choice B, both \(\ell\) and the second line are perpendicular to \(n\), so they are parallel. 3. Choice C has no condition tying the resulting line to \(\ell\).

B

- Separate the direction requirement from the point-of-passage requirement. - Compare how each of the two relevant lines relates to the same third line. - Check the point named in the final construction step.

1. Line \(m\) is perpendicular to \(\ell\), and line \(n\) is also perpendicular to \(m\). 2. Two coplanar lines perpendicular to the same line are parallel, so \(n\parallel\ell\). 3. Line \(n\) passes through \(Q\), not through \(P\), so it does not satisfy the requirement that the parallel line pass through \(P\). 4. In step 3, the perpendicular to \(m\) must be created through \(P\) rather than through \(Q\).

a) Yes. Both \(n\) and \(\ell\) are perpendicular to \(m\), so \(n\parallel\ell\). b) No. It is parallel to \(\ell\), but it does not pass through \(P\). c) Create the second perpendicular through \(P\), not through \(Q\).

- Translate each midpoint or reflection statement into a diagonal relationship. - Identify the quadrilateral whose diagonals meet at \(M\). - Use its classification to obtain the required pair of parallel sides.

1. Since \(M\) is the midpoint of \(\overline{BP}\), \(BM=MP\). 2. Since \(Q\) is the reflection of \(A\) across \(M\), \(M\) is also the midpoint of \(\overline{AQ}\). 3. In quadrilateral \(ABQP\), diagonals \(\overline{AQ}\) and \(\overline{BP}\) bisect each other at \(M\). 4. Therefore, \(ABQP\) is a parallelogram. 5. Opposite sides of a parallelogram are parallel, so \(PQ\parallel AB\). 6. Since \(\overline{AB}\) lies on \(\ell\), \(PQ\parallel\ell\).

The construction makes \(M\) the midpoint of both diagonals \(\overline{AQ}\) and \(\overline{BP}\) of quadrilateral \(ABQP\). The quadrilateral is a parallelogram, so \(PQ\parallel AB\), and therefore \(PQ\parallel\ell\).

- Construct each perpendicular bisector independently from the endpoints of its segment. - Two lines perpendicular to the same line have a fixed relationship. - The distance between the parallel bisectors can be read along \(\ell\) between the two constructed midpoints.

1. For \(\overline{AB}\), draw equal-radius arcs centered at \(A\) and \(B\) with radius greater than half of \(AB\); draw the line through the two arc intersections. This is \(p\). 2. Repeat the same construction from \(B\) and \(C\) to obtain \(q\). 3. Both \(p\) and \(q\) are perpendicular to \(\ell\), so they are parallel. 4. Their intersections with \(\ell\) are the midpoints of \(AB\) and \(BC\), respectively. 5. Those midpoints are \(2.4\,\text{cm}\) on either side of \(B\), so their separation, and hence the perpendicular distance between \(p\) and \(q\), is \(4.8\,\text{cm}\).

Construct each perpendicular bisector with equal-radius arcs from the endpoints of its segment. The two resulting lines are parallel because both are perpendicular to \(\ell\), and their distance is \(4.8\,\text{cm}\).

- Recall the relationship between a chord and the segment joining its midpoint to the center. - Compare that relationship with the perpendicular created at \(O\). - Use a theorem about two lines perpendicular to the same line.

1. Since \(M\) is the midpoint of chord \(\overline{AB}\), the segment from the center to \(M\) is perpendicular to the chord. Thus \(OM\perp AB\). 2. The app constructs \(n\perp OM\). 3. Two distinct lines perpendicular to the same line are parallel. 4. Therefore, \(n\parallel AB\).

The center-to-midpoint segment \(\overline{OM}\) is perpendicular to chord \(\overline{AB}\), and line \(n\) is also perpendicular to \(\overline{OM}\). Therefore, \(n\parallel AB\).

- Identify the transversal and compare the locations of the two equal angles. - Decide which angle-pair relationship supports a converse theorem about parallel lines. - Check how changing sides of the transversal changes the geometric configuration.

1. The original angle at \(A\) and the copied angle at \(P\) lie between lines \(\ell\) and \(n\) on opposite sides of transversal \(\overline{AP}\). 2. The angles are congruent by the copying procedure, so they form a pair of congruent alternate interior angles. 3. By the converse of the alternate interior angles theorem, \(n\parallel\ell\). 4. If the angle is copied on the same side of the transversal as the original angle, the two congruent angles are not alternate interior angles. No parallel-line converse applies, and the resulting line generally intersects \(\ell\).

a) The copied angle and the original angle are congruent alternate interior angles, so the converse theorem gives \(n\parallel\ell\). b) The same-side placement does not create congruent alternate interior angles. It therefore does not justify a parallel line and generally produces a line that intersects \(\ell\).

- Compare the two segments that measure the offset from the original line. - Look for a quadrilateral test involving one pair of opposite sides. - Once the quadrilateral is classified, use its remaining opposite sides.

1. Since \(PF\perp\ell\) and \(GH\perp\ell\), segments \(\overline{PF}\) and \(\overline{GH}\) are parallel. 2. The transferred length gives \(PF=GH\). 3. In quadrilateral \(FGHP\), one pair of opposite sides, \(\overline{PF}\) and \(\overline{GH}\), is both parallel and congruent. 4. Therefore, \(FGHP\) is a parallelogram. 5. Opposite sides of a parallelogram are parallel, so \(PH\parallel FG\). Because \(\overline{FG}\) lies on \(\ell\), \(PH\parallel\ell\).

The equal perpendicular offsets make \(\overline{PF}\) and \(\overline{GH}\) parallel and congruent. Thus \(FGHP\) is a parallelogram, so \(PH\parallel FG\), and therefore \(PH\parallel\ell\).

- How could two triangles made from the same pair of rays have matching angles? - Which measurements should represent \(3:5\), and which should represent \(7:x\)? - Explain the side ratios only after deciding where the parallel segment belongs.

1. Draw two rays from point \(O\). On the first ray, mark \(A\) and \(B\) so that \(OA=3\,\text{cm}\) and \(OB=5\,\text{cm}\). 2. On the second ray, mark \(C\) so that \(OC=7\,\text{cm}\). 3. Join \(A\) to \(C\). Through \(B\), draw a line parallel to \(AC\), meeting the second ray at \(D\). 4. Since \(AC\parallel BD\), \(\triangle OAC\sim\triangle OBD\). Thus, \(\frac{OA}{OB}=\frac{OC}{OD}\). 5. Substitute: \(\frac{3}{5}=\frac{7}{OD}\), so \(OD=\frac{35}{3}\,\text{cm}\). The constructed length \(x\) is \(OD\).

From \(O\), draw two rays. Mark \(OA=3\,\text{cm}\), \(OB=5\,\text{cm}\) on the first and \(OC=7\,\text{cm}\) on the second. Join \(A\) to \(C\), then draw the parallel to \(AC\) through \(B\); it meets the second ray at \(D\). The triangles are similar by AA, so \(\frac{3}{5}=\frac{7}{OD}\). Therefore, \(x=OD=\frac{35}{3}\,\text{cm}\).