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Parallel Lines and Transversals: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two position descriptions

In the numbered diagram, identify the family of each pair. The lines $\ell$ and $m$ are parallel.

A. $\angle1$ and $\angle7$ .

B. $\angle4$ and $\angle5$ .

The numbered sectors at the two crossings.

Text description of this figure

Two parallel horizontal lines, l above m, are crossed by a transversal descending from upper left to lower right. Upper northwest, northeast, southeast and southwest angles are numbered 1, 2, 3 and 4. The lower angles in the same order are 5, 6, 7 and 8. Matching arrow marks give the parallel relationship. No angle measures are displayed.
1. Hint 1
  First decide whether each angle lies between the two crossed lines or outside them.
2. Hint 2
  For each pair, compare the sides of the transversal on which the angles lie.
Answer

A. Alternate exterior. B. Same-side interior.

Full solution

A.
Angles 1 and 7 lie outside the two lines and on opposite sides of the transversal.
They are alternate exterior angles.

B.
Angles 4 and 5 lie between the two lines, both on the left of the transversal.
They are same-side interior angles.

Answer

A. Alternate exterior. B. Same-side interior.

Key idea

An angle family records interior or exterior position and the side of one transversal.
Problem 2 One interior partner

In the numbered diagram, $\ell\parallel m$ and $m\angle6=104^\circ$ . Find $m\angle3$ and justify the relationship you use.

The numbered sectors at the two crossings.

Text description of this figure

Two parallel horizontal lines, l above m, are crossed by a transversal descending from upper left to lower right. Upper northwest, northeast, southeast and southwest angles are numbered 1, 2, 3 and 4. The lower angles in the same order are 5, 6, 7 and 8. Matching arrow marks give the parallel relationship. No angle measures are displayed.
1. Hint 1
  Compare where angles 3 and 6 lie relative to the parallel lines and the transversal.
2. Hint 2
  They are a same-side interior pair. Write the required sum before finding the unknown measure.
Answer

$m\angle3=76^\circ$ .

Full solution

Angles 3 and 6 are interior and on the same side of the transversal.
The given parallelism makes their measures supplementary.

Subtract the known measure from a straight-angle total.
$m\angle3=180^\circ-104^\circ$

Therefore
$m\angle3=76^\circ.$
The check is
$76^\circ+104^\circ=180^\circ.$

Answer

$m\angle3=76^\circ$ .

Key idea

Use the position of a same-side interior pair and the parallel hypothesis to justify a supplementary equation.
Problem 3 Two exterior requests

In the numbered diagram, $\ell\parallel m$ and $m\angle1=41^\circ$ .

A. Find $m\angle7$ and name its relationship to $\angle1$ .

B. Find $m\angle8$ and name its relationship to $\angle1$ .

The numbered sectors at the two crossings.

Text description of this figure

Two parallel horizontal lines, l above m, are crossed by a transversal descending from upper left to lower right. Upper northwest, northeast, southeast and southwest angles are numbered 1, 2, 3 and 4. The lower angles in the same order are 5, 6, 7 and 8. Matching arrow marks give the parallel relationship. No angle measures are displayed.
1. Hint 1
  Both requested angles are exterior. Check whether each is on the same side of the transversal as angle 1.
2. Hint 2
  For parallel lines, an alternate exterior pair is congruent and a same-side exterior pair is supplementary.
Answer

A. $m\angle7=41^\circ$ , alternate exterior. B. $m\angle8=139^\circ$ , same-side exterior.

Full solution

A.
Angles 1 and 7 are alternate exterior angles of parallel lines, so their measures are equal.
Thus
$m\angle7=41^\circ.$

B.
Angles 1 and 8 are same-side exterior angles of parallel lines, so their measures total $180^\circ$ .
$m\angle8=180^\circ-41^\circ$

B.
Hence
$m\angle8=139^\circ.$
At the lower crossing, angles 7 and 8 also form a linear pair, and
$41^\circ+139^\circ=180^\circ.$

Answer

A. $m\angle7=41^\circ$ , alternate exterior. B. $m\angle8=139^\circ$ , same-side exterior.

Key idea

Exterior pairs can require equality or a supplementary sum, depending on their sides of the transversal.
Problem 4 An equality from positions

In the numbered diagram, $\ell\parallel m$ , $m\angle4=(6x+4)^\circ$ , and $m\angle6=(9x-35)^\circ$ . Find $x$ and both angle measures. Justify the equation.

The numbered sectors with a rising transversal.

Text description of this figure

Two parallel horizontal lines l above m are crossed by a transversal rising from lower left to upper right. At the upper crossing the northwest, northeast, southeast and southwest sectors are numbered 1, 2, 3 and 4, and the lower sectors in the same order are 5, 6, 7 and 8. No angle measures or answers are displayed.
1. Hint 1
  Locate the pair before deciding whether to set the expressions equal or add them.
2. Hint 2
  Angles 4 and 6 are alternate interior angles. Use the given parallelism, then solve the resulting linear equation.
3. Hint 3
  After solving, substitute into both original expressions and compare their measures.
Answer

$x=13$ ; $m\angle4=m\angle6=82^\circ$ .

Full solution

Angles 4 and 6 are alternate interior angles of parallel lines.
Their congruence gives equality of measures.
$6x+4=9x-35$

Subtract $6x$ and add $35$ to get $39=3x$ .
Dividing by 3 gives $x=13$ .

Substitution into the first expression gives
$6(13)+4=82.$
The other expression gives
$9(13)-35=82.$
Thus both angles measure $82^\circ$ , which is a valid nonstraight angle measure.

Answer

$x=13$ ; $m\angle4=m\angle6=82^\circ$ .

Key idea

A justified congruence gives an equation of equal measures, and substitution checks the geometry after the algebra.
Problem 5 An exterior sum equation

In the numbered diagram, $\ell\parallel m$ , $m\angle2=(3y+5)^\circ$ , and $m\angle7=(7y+15)^\circ$ . Find $y$ and both angle measures. Justify the equation.

The numbered sectors with a rising transversal.

Text description of this figure

Two parallel horizontal lines l above m are crossed by a transversal rising from lower left to upper right. At the upper crossing the northwest, northeast, southeast and southwest sectors are numbered 1, 2, 3 and 4, and the lower sectors in the same order are 5, 6, 7 and 8. No angle measures or answers are displayed.
1. Hint 1
  Identify whether the two exterior angles lie on the same or opposite sides of the transversal.
2. Hint 2
  A same-side exterior pair of parallel lines is supplementary. Add the two expressions and use that total.
3. Hint 3
  Combine like terms, solve, and substitute into each original measure.
Answer

$y=16$ ; $m\angle2=53^\circ$ and $m\angle7=127^\circ$ .

Full solution

Angles 2 and 7 are same-side exterior angles of parallel lines, so their measures sum to $180^\circ$ .
$(3y+5)+(7y+15)=180$

Combining terms gives
$10y+20=180.$
Subtracting 20 gives $10y=160$ , so $y=16$ .

Then $m\angle2=3(16)+5=53^\circ$ and
$m\angle7=7(16)+15=127^\circ.$
Both are positive and less than $180^\circ$ ; their sum is $180^\circ$ .

Answer

$y=16$ ; $m\angle2=53^\circ$ and $m\angle7=127^\circ$ .

Key idea

The correct geometric sum must be chosen before an angle-expression equation is solved.
Problem 6 Across three crossings

The three-line diagram gives $\ell\parallel m$ and $m\parallel n$ . The same transversal meets all three lines at distinct points, and $m\angle A=128^\circ$ . Find $m\angle B$ and $m\angle C$ , explaining how the information reaches the bottom crossing.

One transversal meets three lines.

Text description of this figure

Three lines l, m and n are drawn across one transversal. The given relations are l parallel to m and m parallel to n. The northeast angle at the upper crossing is A. At the lower crossing the northwest angle is B and the southeast angle is C. No measures are displayed.
1. Hint 1
  Carry a matching sector from the upper line to the middle line, then from the middle line to the lower line.
2. Hint 2
  After transferring the northeast angle twice, use a linear pair and vertical angles at the lower crossing.
Answer

$m\angle B=52^\circ$ and $m\angle C=52^\circ$ .

Full solution

For $\ell\parallel m$ , corresponding congruence carries the upper northeast measure $128^\circ$ to the middle northeast angle.
For $m\parallel n$ , it carries that same middle measure to the lower northeast angle.

Angle B is adjacent to that lower northeast angle in a linear pair.
$m\angle B=180^\circ-128^\circ$

Thus
$m\angle B=52^\circ.$
Angle C is vertical to angle B at the lower crossing, so $m\angle C=52^\circ$ as well.

Answer

$m\angle B=52^\circ$ and $m\angle C=52^\circ$ .

Key idea

Transfer a sector one given parallel pair at a time, then use the angle relationships at the final crossing.
Problem 7 Two routes to one vertex

In the diagram, $\ell\parallel m$ . Transversals meet at $P$ on $\ell$ and meet $m$ at $Q$ and $R$ , with $Q$ to the left of $R$ . The interior angle at $Q$ between $QP$ and the rightward ray of $m$ measures $43^\circ$ . The interior angle at $R$ between $RP$ and the leftward ray of $m$ measures $59^\circ$ . Find $m\angle QPR$ using the parallel lines and angle addition.

Two transversals bring information to P.

Text description of this figure

Lines l and m are parallel and horizontal. Two transversals meet at P on l, then cross m at Q on the left and R on the right. The angle at Q between QP and the rightward ray of m is 43 degrees. The angle at R between RP and the leftward ray of m is 59 degrees. The angle QPR is marked x.
1. Hint 1
  Follow PQ and PR separately to move the two given measures to P.
2. Hint 2
  Alternate interior congruence places one given measure on each side of angle QPR below line l.
3. Hint 3
  The three adjacent angles below l form a straight angle. Subtract the two known parts.
Answer

$m\angle QPR=78^\circ$ .

Full solution

Using transversal PQ, alternate interior congruence gives a $43^\circ$ angle between the leftward ray of $\ell$ and $PQ$ .
Using transversal PR gives a $59^\circ$ angle between $PR$ and the rightward ray of $\ell$ .

Those two angles and $\angle QPR$ are nonoverlapping parts of the straight angle below $\ell$ .
$m\angle QPR=180^\circ-43^\circ-59^\circ$

Hence
$m\angle QPR=78^\circ.$
The check is
$43^\circ+78^\circ+59^\circ=180^\circ.$

Answer

$m\angle QPR=78^\circ$ .

Key idea

Two transversals can transfer different measures to adjacent parts of the same straight angle.
Problem 8 Complete the connecting reason

In the numbered diagram, $\ell\parallel m$ . An argument establishes $m\angle1+m\angle4=180^\circ$ from a linear pair, and $m\angle4+m\angle5=180^\circ$ from the same-side interior theorem.

State the equality that follows by subtracting the same measure from these sums. Explain why it establishes a corresponding-angle congruence without assuming that congruence as a postulate.

The numbered sectors at the two crossings.

Text description of this figure

Two parallel horizontal lines, l above m, are crossed by a transversal descending from upper left to lower right. Upper northwest, northeast, southeast and southwest angles are numbered 1, 2, 3 and 4. The lower angles in the same order are 5, 6, 7 and 8. Matching arrow marks give the parallel relationship. No angle measures are displayed.
1. Hint 1
  Find the measure that occurs in both sums.
2. Hint 2
  The two sums equal the same total. Subtract the shared measure and interpret the remaining equality as angle congruence.
Answer

$m\angle1=m\angle5$ , so $\angle1\cong\angle5$ is a corresponding-angle congruence.

Full solution

Both sums equal $180^\circ$ .
Subtracting the common $m\angle4$ from each leaves
$m\angle1=m\angle5$

Equal measures establish $\angle1\cong\angle5$ .
These angles occupy matching northwest sectors at the two crossings, so they are corresponding.

The interior sum was proved from the accepted angle postulate, and the other sum is a local linear-pair result.
Their common supplement therefore derives the congruence rather than assuming it.

Answer

$m\angle1=m\angle5$ , so $\angle1\cong\angle5$ is a corresponding-angle congruence.

Key idea

Corresponding congruence follows from the proved interior sum and a local linear pair by subtracting a common measure.
Problem 9 A total is not the whole check

A transversal crosses two given distinct parallel lines. Let L be the sum of the two interior angles on its left and R the sum on its right. Compare these proposed records.

A. $L=180^\circ$ and $R=180^\circ$ .

B. $L=186^\circ$ and $R=174^\circ$ .

Which record is ruled out by the angle form of Euclid's parallel postulate, and why? State what this check says about the other record, without reversing an angle theorem.
1. Hint 1
  Check each side separately, not only the combined total.
2. Hint 2
  The angle postulate predicts an intersection when one same-side interior sum is below 180 degrees.
3. Hint 3
  A record agreeing with a necessary condition is not, by that fact alone, a proof of parallelism.
Answer

B is ruled out because its right sum is below $180^\circ$ ; A is not ruled out by this check.

Full solution

A.
Both proposed sums equal $180^\circ$ , so neither triggers the postulate's below- $180^\circ$ condition.
The record agrees with the necessary sums for the given parallel lines.
This check does not reverse the theorem to prove parallelism from a record.

B.
The total still equals $360^\circ$ , but $R=174^\circ$ is below $180^\circ$ .
The angle postulate forces the crossed lines to meet on the right side.
That contradicts the given parallelism, so record B is ruled out.

Answer

B is ruled out because its right sum is below $180^\circ$ ; A is not ruled out by this check.

Key idea

The 360-degree combined total and the prohibition of either side being below 180 degrees are both needed.
Problem 10 Keep and release the parallel condition

Use the numbered sector convention: upper northwest, northeast, southeast, southwest are $1,2,3,4$ ; lower sectors in the same order are $5,6,7,8$ . A transversal crosses two distinct coplanar lines at distinct points.

A. The lines are parallel and $m\angle1=58^\circ$ . Find $m\angle5$ and $m\angle2$ .

B. Remove the parallel condition but retain $m\angle1=58^\circ$ . Can $m\angle5=79^\circ$ now occur? If so, give $m\angle7$ and $m\angle6$ , check the local angle rules, and identify the conclusion from A that is no longer forced.
1. Hint 1
  Separate a fact within one crossing from a fact that connects the two crossings.
2. Hint 2
  In A, corresponding congruence connects angles 1 and 5. A local linear pair determines angle 2.
3. Hint 3
  In B, start with the proposed lower measure and apply vertical angles and a linear pair at that lower crossing. Compare it with the still-valid upper crossing.
Answer

A. $m\angle5=58^\circ$ and $m\angle2=122^\circ$ . B. Yes; $m\angle7=79^\circ$ and $m\angle6=101^\circ$ . The conclusion $m\angle5=58^\circ$ is no longer forced; the local vertical and linear-pair rules still hold.

Full solution

A.
Parallelism gives corresponding congruence, so
$m\angle5=58^\circ.$
Angles 1 and 2 are a linear pair.
$m\angle2=180^\circ-58^\circ$

A.
Thus
$m\angle2=122^\circ.$
That step uses only the upper crossing, while the determination of angle 5 uses parallelism.

B.
Yes.
The solution diagram gives a nonparallel arrangement with upper northwest angle $58^\circ$ and lower northwest angle $79^\circ$ .
At the lower crossing, vertical angles give
$m\angle7=79^\circ.$
A linear pair gives
$m\angle6=180^\circ-79^\circ$

B.
Thus $m\angle6=101^\circ$ , and its vertical partner angle 8 also measures $101^\circ$ .
At the upper crossing, the opposite pair is $58^\circ$ each and the other pair $122^\circ$ each.
Every local neighboring pair totals $180^\circ$ , and every local opposite pair agrees.

B.
Angles 1 and 5 can now differ, so the earlier conclusion $m\angle5=58^\circ$ is no longer forced.
The upper linear-pair conclusion $m\angle2=122^\circ$ survives because it never depended on parallelism.

Case B satisfies the local angle rules while the crossed lines are not parallel.

Answer

A. $m\angle5=58^\circ$ and $m\angle2=122^\circ$ . B. Yes; $m\angle7=79^\circ$ and $m\angle6=101^\circ$ . The conclusion $m\angle5=58^\circ$ is no longer forced; the local vertical and linear-pair rules still hold.

Key idea

Removing parallelism releases the connection between crossings while preserving the angle facts at each individual crossing.

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