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Proving Lines Parallel or Perpendicular: 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 Name the sufficient test

Distinct coplanar lines $\ell,m$ are cut by transversal $t$ at distinct points. The diagram numbers the upper sectors northwest, northeast, southeast, southwest as 1, 2, 3, 4 and the lower sectors in the same order as 5, 6, 7, 8.

Case A. $m\angle2=m\angle6=35^\circ$ .

Case B. $m\angle4=75^\circ$ and $m\angle5=105^\circ$ .

For each independent case, decide whether $\ell\parallel m$ follows and name the converse that justifies your decision.

The numbered sectors identify the pairs in both cases.

Text description of this figure

The transversal rises from lower left to upper right. At the upper crossing, sectors 1, 2, 3, 4 occupy northwest, northeast, southeast, southwest positions. The lower crossing has 5, 6, 7, 8 in the same order. No parallel markings or derived measures are shown.
1. Hint 1
  Locate each pair relative to the two lines and transversal.
2. Hint 2
  A corresponding pair needs equal measures; a same-side interior pair needs a total of $180^\circ$ .
Answer

Case A: $\ell\parallel m$ , corresponding-angle converse. Case B: $\ell\parallel m$ , same-side interior converse.

Full solution

Case A.
Angles 2 and 6 are corresponding.
Their stated equal measures make them congruent, so the corresponding-angle converse gives $\ell\parallel m$ .

Case B.
Angles 4 and 5 are same-side interior angles.
Their measures add to $75^\circ+105^\circ=180^\circ$ , so the same-side interior converse gives $\ell\parallel m$ .

Answer

Case A: $\ell\parallel m$ , corresponding-angle converse. Case B: $\ell\parallel m$ , same-side interior converse.

Key idea

A parallelism test needs both the right positions and the required measure relationship.
Problem 2 Make an alternate pair agree

Distinct coplanar lines $\ell,m$ are cut by transversal $t$ at distinct points. The diagram numbers the upper sectors northwest, northeast, southeast, southwest as 1, 2, 3, 4 and the lower sectors in the same order as 5, 6, 7, 8.

The measures are $m\angle3=(7u-13)^\circ$ and $m\angle5=(4u+44)^\circ$ . Find $u$ that establishes $\ell\parallel m$ , find both angle measures, and name the reason for parallelism.

Use the stated expressions for sectors 3 and 5.

Text description of this figure

The transversal rises from lower left to upper right. At the upper crossing, sectors 1, 2, 3, 4 occupy northwest, northeast, southeast, southwest positions. The lower crossing has 5, 6, 7, 8 in the same order. No parallel markings or derived measures are shown.
1. Hint 1
  Decide what relationship an alternate interior pair must satisfy.
2. Hint 2
  Equate the two expressions before using a converse.
3. Hint 3
  After solving, substitute into both expressions and check that each is a valid angle measure.
Answer

$u=19$ ; $m\angle3=m\angle5=120^\circ$ ; alternate interior converse.

Full solution

Angles 3 and 5 are alternate interior angles, so a sufficient condition is equality of their measures.
$7u-13=4u+44$

Subtract $4u$ and add 13 to obtain $3u=57$ .
Dividing by 3 gives $u=19$ .

Substitution gives $7(19)-13=120$ and
$4(19)+44=120.$
Both are valid non-reflex measures.
The alternate interior converse establishes $\ell\parallel m$ .

Answer

$u=19$ ; $m\angle3=m\angle5=120^\circ$ ; alternate interior converse.

Key idea

Solve for the angle condition, verify it, and then cite the converse.
Problem 3 Supplement an exterior pair

Distinct coplanar lines $\ell,m$ are cut by transversal $t$ at distinct points. The diagram numbers the upper sectors northwest, northeast, southeast, southwest as 1, 2, 3, 4 and the lower sectors in the same order as 5, 6, 7, 8.

Angles 1 and 8 measure $(6v+8)^\circ$ and $(3v+19)^\circ$ . Find the value of $v$ that establishes $\ell\parallel m$ , and find the two measures.

Sectors 1 and 8 form the selected exterior pair.

Text description of this figure

The transversal rises from lower left to upper right. At the upper crossing, sectors 1, 2, 3, 4 occupy northwest, northeast, southeast, southwest positions. The lower crossing has 5, 6, 7, 8 in the same order. No parallel markings or derived measures are shown.
1. Hint 1
  These exterior angles lie on the same side of the transversal.
2. Hint 2
  Set their sum, rather than their difference, equal to the measure of a straight angle.
3. Hint 3
  Combine like terms, then check the two measures by substitution.
Answer

$v=17$ ; $m\angle1=110^\circ$ and $m\angle8=70^\circ$ .

Full solution

Angles 1 and 8 are same-side exterior angles.
The converse requires their sum to be $180^\circ$ .
$(6v+8)+(3v+19)=180$

Combining terms gives $9v+27=180$ .
Subtracting 27 and dividing by 9 gives $v=17$ .

The measures become $6(17)+8=110^\circ$ and
$3(17)+19=70^\circ.$
Their sum is $180^\circ$ , so the same-side exterior converse gives $\ell\parallel m$ .

Answer

$v=17$ ; $m\angle1=110^\circ$ and $m\angle8=70^\circ$ .

Key idea

Same-side angles establish parallelism through supplementarity, not congruence.
Problem 4 The second crossing

Distinct coplanar lines $a$ and $b$ are parallel. Transversal $t$ is perpendicular to $a$ . At the crossing of $t$ and $b$ , one angle measures $(11w+2)^\circ$ . Find $w$ and state the relationship between $t$ and $b$ , with a reason.
1. Hint 1
  Relate the crossing at $b$ to the known right angle at $a$ .
2. Hint 2
  The perpendicular transversal theorem supplies the angle measure needed for the equation.
Answer

$w=8$ and $t\perp b$ , by the perpendicular transversal theorem.

Full solution

Because $a\parallel b$ and $t\perp a$ in one plane, the perpendicular transversal theorem gives $t\perp b$ .
The stated angle is therefore right.
$11w+2=90$

Subtract 2 to get $11w=88$ , then divide by 11 to obtain $w=8$ .
Substitution checks
$11(8)+2=90.$

Answer

$w=8$ and $t\perp b$ , by the perpendicular transversal theorem.

Key idea

A perpendicular transversal to one of two parallel lines is perpendicular to both.
Problem 5 Measure a sign on a plan

A scale drawing represents a sign as point $P$ and the edge of a straight walkway as line $m$ , in one plane. Points $H,K$ lie on $m$ , $PH\perp m$ , $PH=26$ meters, and $PK=31$ meters.

Name the foot of the perpendicular, give the distance from the sign to the walkway line, and state which segment you would mark to record that distance.

The right-angle mark is part of the supplied information.

Text description of this figure

Point P lies above line m. H and K lie on m; PH has a right-angle mark at H and is labeled 26 meters. The oblique segment PK is labeled 31 meters.
1. Hint 1
  Use the perpendicular mark to identify the foot.
2. Hint 2
  The distance definition uses the perpendicular joining segment.
Answer

Foot: $H$ . Distance: $26$ meters. Mark segment $\overline{PH}$ .

Full solution

The segment from $P$ meets line $m$ at a right angle at $H$ , so $H$ is the foot of the perpendicular.

Point-to-line distance is defined by the perpendicular segment.
Its length is $PH=26$ meters, so mark $\overline{PH}$ .
The oblique segment $PK$ does not define that distance.

Answer

Foot: $H$ . Distance: $26$ meters. Mark segment $\overline{PH}$ .

Key idea

The perpendicular foot identifies the segment used for point-to-line distance.
Problem 6 Repair a parallelism claim

In one plane, distinct lines $\ell$ and $t$ cross at $P$ . Angles 2 and 4 are vertical angles there, each measuring $39^\circ$ . A different line $m$ crosses $t$ at another point.

Inez writes, “The congruent angles show that $\ell\parallel m$ .” Explain why the information does not justify that conclusion. Sketch one possible choice of $m$ that preserves the stated angles and is not parallel to $\ell$ .
1. Hint 1
  Ask which intersections the given angles describe.
2. Hint 2
  Could the direction of $m$ change while the crossing at $P$ stays fixed?
Answer

The conclusion is unjustified. One valid sketch chooses $S$ on $\ell$ and $R$ on $t$ , both different from $P$ , and draws $m=\overleftrightarrow{RS}$ .

Full solution

The given congruent angles are vertical angles at $P$ .
The vertical-angle theorem supplies their equality without using $m$ , so the equality does not restrict the direction of $m$ .

Choose $S$ on $\ell$ different from $P$ and $R$ on $t$ different from $P$ .
Draw
$m=\overleftrightarrow{RS}.$
This line crosses $t$ at $R$ and meets $\ell$ at $S$ , so it is not parallel to $\ell$ .
The crossing at $P$ and both $39^\circ$ vertical-angle givens remain unchanged.

One possible second line preserves the local angle givens while failing the parallel conclusion.

Answer

The conclusion is unjustified. One valid sketch chooses $S$ on $\ell$ and $R$ on $t$ , both different from $P$ , and draws $m=\overleftrightarrow{RS}$ .

Key idea

A fact at one crossing cannot by itself test the direction of a second line.
Problem 7 Perform and annotate the construction

On paper, draw the given line $\ell$ and an external point $P$ in the same plane. Using a compass and unmarked straightedge, construct a line $m$ through $P$ parallel to $\ell$ .

Choose and label distinct points $A,B$ on $\ell$ . Use an auxiliary line through $A,P$ and label $C$ beyond $P$ so the order is $A,P,C$ . Retain the angle-copy arcs, label a point $Q$ on the constructed ray, mark the copied angles with matching arcs, and write a justification of parallelism.

Begin with the supplied line and external point.

Text description of this figure

Only line ℓ and point P outside it are given in one plane. No auxiliary line, copied ray, compass arc, or final line has been drawn.
1. Hint 1
  Connect the outside point to a point of the given line to create a transversal.
2. Hint 2
  Copy $\angle PAB$ using ray $PC$ as the starting side, placing the new ray on the same side of $AP$ as $B$ .
3. Hint 3
  Identify the two copied angles by their positions relative to the transversal.
Answer

A constructed line $m=\overleftrightarrow{PQ}$ through $P$ with $m\parallel\ell$ , with matching copy arcs for $\angle CPQ$ and $\angle PAB$ .

Full solution

Choose distinct $A,B$ on $\ell$ and draw $AP$ .
Extend beyond $P$ to place $C$ , giving the order $A,P,C$ .

Use the taught angle-copy procedure to copy $\angle PAB$ at $P$ from starting ray $PC$ , on the same side of $AP$ as $B$ .
Retain the equal-radius arcs and the chord-transfer arc.
Draw the new ray through its transferred point $Q$ .

The angle-copy procedure gives $\angle CPQ\cong\angle PAB$ .
Its correctness is proved later in lesson 4.5; that copying fact is accepted here.
Extend ray $PQ$ to line $m$ .

The copied angles are corresponding for transversal $AP$ .
The crossings are distinct because $P$ is outside $\ell$ , so the corresponding-angle converse establishes $m\parallel\ell$ .
Matching arcs record the copied-angle given; parallel arrows may now record the established conclusion.

A possible completed construction with the copying record retained.

Answer

A constructed line $m=\overleftrightarrow{PQ}$ through $P$ with $m\parallel\ell$ , with matching copy arcs for $\angle CPQ$ and $\angle PAB$ .

Key idea

A construction needs both the copying record and a valid converse connecting it to parallelism.
Problem 8 Use uniqueness at an assumed crossing

Three pairwise distinct coplanar lines satisfy $a\parallel c$ and $b\parallel c$ . An indirect proof temporarily assumes that $a$ and $b$ meet at $R$ .

Explain why $R$ is outside $c$ , identify the contradiction, and classify $a$ and $b$ .
1. Hint 1
  A point on $a$ cannot also lie on $c$ under the given parallelism.
2. Hint 2
  Count the claimed parallels to $c$ through the assumed point $R$ .
Answer

$R$ would be outside $c$ ; two distinct parallels to $c$ through $R$ contradict uniqueness. Therefore $a\parallel b$ .

Full solution

Because $a\parallel c$ , no point of $a$ lies on $c$ .
The assumed common point $R$ lies on $a$ , so it is outside $c$ .

Both $a$ and $b$ would pass through $R$ and be parallel to $c$ .
They are given distinct, contradicting the accepted uniqueness of a parallel through an external point.

Reject the assumed crossing.
The distinct lines $a$ and $b$ lie in one plane and do not meet, so they are parallel.

Answer

$R$ would be outside $c$ ; two distinct parallels to $c$ through $R$ contradict uniqueness. Therefore $a\parallel b$ .

Key idea

An indirect proof must name the exact accepted fact contradicted by its temporary assumption.
Problem 9 Two conditions on one pair

Distinct coplanar lines $\ell,m$ are cut by transversal $t$ at distinct points. The diagram numbers the upper sectors northwest, northeast, southeast, southwest as 1, 2, 3, 4 and the lower sectors in the same order as 5, 6, 7, 8.

Angles 4 and 5 are congruent and supplementary. Find each measure, classify the relationship between $\ell$ and $m$ , and classify the relationship of $t$ to each. On a copy of the diagram, add only markings justified by your conclusions.

No conclusion markings are supplied.

Text description of this figure

The transversal rises from lower left to upper right. At the upper crossing, sectors 1, 2, 3, 4 occupy northwest, northeast, southeast, southwest positions. The lower crossing has 5, 6, 7, 8 in the same order. No parallel markings or derived measures are shown.
1. Hint 1
  Equal measures with a fixed total determine both measures.
2. Hint 2
  Use supplementarity for the parallelism test, then interpret the individual measures.
Answer

$m\angle4=m\angle5=90^\circ$ ; $\ell\parallel m$ ; $t\perp\ell$ and $t\perp m$ . Add matching parallel arrows and right-angle marks at both crossings.

Full solution

Let each equal measure be $r$ degrees.
Supplementarity gives $r+r=180$ , so $2r=180$ and $r=90$ .
Thus both selected angles are right.

Angles 4 and 5 are same-side interior angles and are supplementary, so their converse gives $\ell\parallel m$ .

The $90^\circ$ angles establish $t\perp\ell$ and $t\perp m$ .
Add right-angle marks at the two crossings and matching parallel arrows on the two lines only after these facts have been justified.

The annotations record conclusions after the reasoning.

Answer

$m\angle4=m\angle5=90^\circ$ ; $\ell\parallel m$ ; $t\perp\ell$ and $t\perp m$ . Add matching parallel arrows and right-angle marks at both crossings.

Key idea

Congruence and supplementarity together force right angles, while the pair positions justify parallelism.
Problem 10 Choose and complete sufficient givens

Distinct coplanar lines $\ell,m$ are cut by transversal $t$ at distinct points. The diagram numbers the upper sectors northwest, northeast, southeast, southwest as 1, 2, 3, 4 and the lower sectors in the same order as 5, 6, 7, 8.

Each packet below is independent. Decide whether it guarantees $\ell\parallel m$ . For each insufficient packet, supply one possible additional angle or perpendicular fact that would make the conclusion follow, and name the valid test.

Packet A. $m\angle1=m\angle3=51^\circ$ .

Packet B. $m\angle4=116^\circ$ and $m\angle5=64^\circ$ .

Packet C. $\ell\perp t$ .

Use the same numbered positions for the three independent packets.

Text description of this figure

The transversal rises from lower left to upper right. At the upper crossing, sectors 1, 2, 3, 4 occupy northwest, northeast, southeast, southwest positions. The lower crossing has 5, 6, 7, 8 in the same order. No parallel markings or derived measures are shown.
1. Hint 1
  A useful packet must constrain the directions at both crossings.
2. Hint 2
  For an incomplete packet, choose a fact about the other crossing that matches a proved converse.
3. Hint 3
  One possible repair may use a corresponding angle, and another may make both crossings right.
Answer

A: insufficient; add $m\angle5=51^\circ$ (corresponding). B: sufficient (same-side interior). C: insufficient; add $m\perp t$ (two perpendiculars). Other valid sufficient additions accepted.

Full solution

Packet A.
Angles 1 and 3 are vertical at one crossing, so their equality does not determine $m$ .
Add
$m\angle5=51^\circ.$
Angles 1 and 5 then form a congruent corresponding pair, and their converse proves parallelism.

Packet B.
Angles 4 and 5 are same-side interior angles.
Their measures sum to $116^\circ+64^\circ=180^\circ$ , so this packet already proves parallelism by the same-side interior converse.

Packet C.
A right angle at $\ell$ alone does not determine the direction of $m$ .
Add $m\perp t$ .
The two distinct coplanar lines then make congruent corresponding right angles with $t$ , proving parallelism.

Other added facts are valid if, together with that packet and the stated incidence conditions, they establish a proved converse or the coplanar two-perpendiculars result.

Answer

A: insufficient; add $m\angle5=51^\circ$ (corresponding). B: sufficient (same-side interior). C: insufficient; add $m\perp t$ (two perpendiculars). Other valid sufficient additions accepted.

Key idea

Choose sufficient givens by tracing an explicit route from their angle positions to a proved parallelism test.

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