Proving lines parallel
Two lines run across a figure and look as if they never meet. Which angles are enough to prove that they never will?
Where you are 18 in Geometric proof
When you reach for it
Whenever the Prove says two lines are parallel. The definition is no help there: “they never meet” is a claim about points as far off as you like, and no figure can show them all. What a figure does offer is angles, eight of them where a transversal crosses the two lines. The facts in Parallel lines and transversals run from parallel lines to angles. This lesson runs them backwards, as converses: from the right pair of angles to parallel lines.
It is the last step of every proof that a quadrilateral is a trapezoid or a parallelogram, and the reason the compass construction of a parallel line works.
The procedure
Number the angles as in Figure 1: at each crossing upper left, upper right, lower left, lower right, so \angle 1 to \angle 4 at l and \angle 5 to \angle 8 at m.
The five tests. The last line of the proof is l \parallel m, and it stands on one of these.
| The pair | proves l \parallel m when it is |
|---|---|
| corresponding, such as \angle 2 and \angle 6 | congruent |
| alternate interior, such as \angle 3 and \angle 6 | congruent |
| alternate exterior, such as \angle 1 and \angle 8 | congruent |
| same side interior, such as \angle 4 and \angle 6 | supplementary |
| the angles t makes with l and with m | both right |
The first four are cited as “converse of the Corresponding Angles Theorem” and so on, one for each pair. The fifth is cited as “two lines perpendicular to one line are parallel”.
Where the tests come from. The first is proved in a theorem and its converse: if l and m met, the two crossings and the meeting point would make a triangle with two angles adding to 180^\circ, and Euclid’s Proposition 17 says no triangle has that. Indirect proof reaches the same test through Proposition 16, the exterior angle inequality. Each other test trades its pair for a corresponding pair, using only facts true at any crossing:
- alternate interior: \angle 3 is vertical to \angle 2, so \angle 3 \cong \angle 6 gives \angle 2 \cong \angle 6;
- alternate exterior: \angle 8 is vertical to \angle 5, so \angle 1 \cong \angle 8 gives \angle 1 \cong \angle 5;
- same side interior: a linear pair, set out in full in Example 1;
- perpendiculars: in Figure 2 the two right angles sit in the same position, and the Right Angle Congruence Theorem makes them congruent, set out in full in Example 5.
None of this uses the parallel postulate. The forward theorems do: on this map the postulate proves the Corresponding Angles Theorem, which proves the Alternate Interior Angles Theorem. Euclid also proved the converses first: Book I, Proposition 27 is the alternate interior test, Proposition 28 the corresponding and same side interior tests, and his postulate is used for the first time in Proposition 29, the forward theorems.
Chase with crossing facts only. When the given angles are not a tested pair, move one of them with vertical angles or linear pairs until they are. Never move an angle with a forward theorem such as “corresponding angles are congruent”: its if-part is l \parallel m, which is the Prove.
Name the transversal, then the lines. Each angle of the pair has one side on the transversal. Its other side lies on one of the two lines the test makes parallel (Example 2).
Cite the converse. The last reason is the converse, whose if-part is the angle fact on the lines above it. A pair that fails its test proves the opposite: if l \parallel m, the forward theorem would force the pair to pass, so by the contrapositive, l \nparallel m.
Worked examples
The same side interior test
Prove the same side interior test from the corresponding angles test.
Given. A transversal t crosses l at P and m at Q; \angle 4 and \angle 6 are supplementary.
Prove. l \parallel m.
Plan. \angle 2 corresponds to \angle 6, and it forms a linear pair with \angle 4. So \angle 2 and \angle 6 are both supplements of \angle 4.
| Statements | Reasons |
|---|---|
| 1. \angle 4 and \angle 6 are supplementary | given |
| 2. m\angle 4 + m\angle 6 = 180^\circ | definition of supplementary angles: line 1 |
| 3. \angle 2 and \angle 4 form a linear pair | definition of linear pair: in Figure 3 they share the right-hand ray of l, and their other sides are opposite rays of t |
| 4. m\angle 2 + m\angle 4 = 180^\circ | Linear Pair Theorem: line 3 |
| 5. 180^\circ = m\angle 4 + m\angle 6 | symmetric property of equality: line 2 |
| 6. m\angle 2 + m\angle 4 = m\angle 4 + m\angle 6 | substitution: line 5 into line 4 |
| 7. m\angle 2 = m\angle 6 | subtraction property of equality: line 6 |
| 8. \angle 2 \cong \angle 6 | definition of congruent angles: line 7 |
| 9. l \parallel m | converse of the Corresponding Angles Theorem: line 8, both angles above their line and right of t |
Figure 3 shades the given pair and marks nothing else, since supplementary has no mark of its own. Line 3 reads positions from the figure, which a diagram may supply. No line before the last mentions parallel lines.
Which lines are parallel?
In quadrilateral ABCD of Figure 4, the diagonal \overline{BD} is drawn. Which two sides are parallel (a) if \angle ABD \cong \angle CDB; (b) if instead \angle DAB and \angle ADC are supplementary?
(a) Both angles have a side on \overline{BD}, so \overleftrightarrow{BD} is the transversal. Their other sides lie on \overleftrightarrow{AB} and \overleftrightarrow{DC}, the two lines it crosses. Both angles open toward the other line, and A and C are on opposite sides of \overline{BD}, so the pair is alternate interior. By its test, \overline{AB} \parallel \overline{DC}.
The pair says nothing about \overline{AD} and \overline{BC}, and in Figure 4 they lean different ways. The pair along \overline{BD} that would test them is \angle ADB and \angle CBD.
(b) The two angles share the side \overline{AD}, so now \overleftrightarrow{AD} is the transversal. Their other sides lie on \overleftrightarrow{AB} and \overleftrightarrow{DC} again, with B and C on the same side of \overline{AD}: a supplementary same side interior pair. Again \overline{AB} \parallel \overline{DC}, and again nothing about the other two sides. Either given proves ABCD a trapezoid. Neither proves it a parallelogram: that takes a second pair, and Figure 4 shows a quadrilateral where the second pair fails.
A pair that proves nothing
Lines l and m are cut by t, numbered as in Figure 1, and \angle 3 \cong \angle 5. Does this prove l \parallel m?
SolutionNo. \angle 3 and \angle 5 are both between the lines and both left of t, a same side interior pair, and its test asks for supplementary angles, not congruent ones.
In Figure 5 both measure 70^\circ. If l \parallel m, the Same Side Interior Angles Theorem would make them add to 180^\circ. They add to 70^\circ + 70^\circ = 140^\circ, so l \nparallel m. The lines close up on the left, where the pair falls short, and meet at R; triangle PQR has a third angle of 180^\circ - 140^\circ = 40^\circ.
There is one exception. Congruent and supplementary together force 90^\circ each, and then t is perpendicular to both lines and the fifth test applies. So a congruent same side pair proves l \parallel m only when both angles are right.
The value that makes the lines parallel
Numbered as in Figure 1, m\angle 1 = (3x + 16)^\circ and m\angle 6 = (5x - 4)^\circ. For what value of x is l \parallel m?
Solution\angle 1 and \angle 6 are not a tested pair, so chase one with a crossing fact. \angle 1 is vertical to \angle 4, so m\angle 4 = (3x + 16)^\circ, and \angle 4 and \angle 6 are a same side interior pair. The test wants them supplementary:
Then m\angle 1 = 3(21) + 16 = 79^\circ and m\angle 6 = 5(21) - 4 = 101^\circ, and 79 + 101 = 180, so by the converse l \parallel m.
Check by another route. \angle 5 forms a linear pair with \angle 6, so m\angle 5 = 180^\circ - 101^\circ = 79^\circ = m\angle 1: a congruent corresponding pair. For any other x the pair is not supplementary, and the lines are not parallel.
The perpendicular test, line by line
Prove the fifth test: two lines perpendicular to one transversal are parallel.
Given. l \perp t at P; m \perp t at Q.
Prove. l \parallel m.
SolutionPlan. Figure 2 is this Given. Perpendicular lines make four right angles, so the angle above each line and right of t, \angle 2 and \angle 6 in the numbering of Figure 1, is right. Right angles are congruent, and these two are corresponding.
| Statements | Reasons |
|---|---|
| 1. l \perp t at P | given |
| 2. m \perp t at Q | given |
| 3. \angle 2 is a right angle | perpendicular lines make four right angles: line 1 |
| 4. \angle 6 is a right angle | perpendicular lines make four right angles: line 2 |
| 5. \angle 2 \cong \angle 6 | Right Angle Congruence Theorem: lines 3 and 4 |
| 6. l \parallel m | converse of the Corresponding Angles Theorem: line 5, both angles above their line and right of t |
The Given measured no angle at all; the Right Angle Congruence Theorem built the congruent pair the test needed. Once this proof is on the page, “two lines perpendicular to one line are parallel” may stand as a single reason, as the table of tests says. With numbers: if a proof shows m\angle 2 = 90^\circ and m\angle 6 = 90^\circ, the test applies; if m\angle 6 = 89^\circ, the corresponding pair fails and l \nparallel m.
A bend that adds up
Given. A is on l, B is on m, and X lies between the lines, to the right of A and B; the angle between l and \overline{AX} at A is 38^\circ, the angle between m and \overline{BX} at B is 47^\circ, and m\angle AXB = 85^\circ (Figure 6, left).
Prove. l \parallel m.
Plan. This is the converse of the bend in Alternate and same side angles, Example 5, where parallel lines made the angle at X the sum of the angles at A and B. No line in the figure has known angles at both l and m, so build one between them. At X, copy the 38^\circ angle to get a line n \parallel l. What is left of the 85^\circ is 47^\circ, which makes n \parallel m.
| Statements | Reasons |
|---|---|
| 1. The angle between l and \overline{AX} at A is 38^\circ | given |
| 2. The angle between m and \overline{BX} at B is 47^\circ | given |
| 3. m\angle AXB = 85^\circ | given |
| 4. m\angle AXN = 38^\circ, with N on the same side of \overleftrightarrow{XA} as B | protractor postulate: exactly one such ray makes 38^\circ |
| 5. \overrightarrow{XN} is inside \angle AXB | protractor postulate: 38 < 85, lines 3 and 4 |
| 6. l \parallel n, where n = \overleftrightarrow{XN} | converse of the Alternate Interior Angles Theorem: lines 1 and 4, transversal \overleftrightarrow{AX} |
| 7. m\angle AXN + m\angle NXB = m\angle AXB | angle addition postulate: line 5 |
| 8. 38^\circ + m\angle NXB = 85^\circ | substitution: lines 4 and 3 into line 7 |
| 9. m\angle NXB = 47^\circ | subtraction property of equality: line 8 |
| 10. n \parallel m | converse of the Alternate Interior Angles Theorem: lines 2 and 9, transversal \overleftrightarrow{BX} |
| 11. l \parallel m | two lines parallel to the same line are parallel (The parallel postulate, Example 4): lines 6 and 10 |
X is on neither l nor m, so n differs from both. Lines 6 and 10 need no postulate; line 11 does, since that lesson proves it from the postulate. All the numbers had to do was satisfy 85 - 38 = 47.
Boundaries
The tests hold in a plane. In space two lines can miss each other without being parallel (skew lines), and the perpendicular test fails: two edges of a box that each meet a third edge at a right angle can be skew. Every test assumes l, m and t lie in one plane.
One transversal, the same two lines. Congruent angles on different transversals, or along a transversal of some other pair of lines, prove nothing about l and m (Example 2).
Congruent is not always the condition. Same side pairs must be supplementary, and a congruent same side pair proves parallel only when both angles are right (Example 3).
Exactly, not nearly. A same side pair of 179.9^\circ proves the lines are not parallel.
A third line brings the postulate in. The tests hold even in geometries where the parallel postulate fails. Joining l \parallel n and n \parallel m into l \parallel m, as Example 6 did, does not: without the postulate, two different lines through one point can both be parallel to a third line, and then they are parallel to the same line and still meet.
Where this goes next
Proofs with parallel lines uses these tests beside the forward theorems, often in one proof, as Exercise 15 does. Two tests at once make a quadrilateral a parallelogram, in proving a quadrilateral is a parallelogram. The construction of a parallel line copies an angle so that a corresponding pair is congruent, and the first test is its proof. And each test, joined to its forward theorem, is a Biconditional: l \parallel m if and only if a corresponding pair is congruent.
Exercises
Which test?
Use the numbering of Figure 1. Say which test, if any, proves l \parallel m from the fact given.
- \angle 1 \cong \angle 5
- \angle 1 \cong \angle 8
- \angle 4 and \angle 6 are supplementary.
- \angle 2 \cong \angle 3
- \angle 2 and \angle 8 are supplementary.
- \angle 2 \cong \angle 8
Parallel or not?
Numbered as in Figure 1.
- m\angle 3 = 58^\circ and m\angle 8 = 122^\circ. Is l \parallel m?
- m\angle 1 = 104^\circ and m\angle 6 = 86^\circ. Is l \parallel m? If not, on which side of t do the lines meet, and at what angle?
- In a figure like Figure 6, the angles at A and B are 31^\circ and 52^\circ. Is l \parallel m when m\angle AXB = 83^\circ? When m\angle AXB = 85^\circ?
Finding the value
- Numbered as in Figure 1, m\angle 2 = (5x - 14)^\circ and m\angle 7 = (3x + 26)^\circ. Find the value of x that makes l \parallel m, and the two angles then.
- m\angle 4 = (7x + 4)^\circ and m\angle 6 = (2x + 41)^\circ. Find x for l \parallel m, and both angles.
- m\angle 3 = (x^2)^\circ and m\angle 5 = (6x + 20)^\circ. Find every x that makes l \parallel m in a real figure, and the angles.
- In quadrilateral ABCD, m\angle A = (3y + 10)^\circ and m\angle B = (5y + 10)^\circ. For what y is \overline{AD} \parallel \overline{BC}, and which side is the transversal?
Writing the proof
Exercise 15 uses Figure 7.
- Write a two-column proof: given \angle 4 \cong \angle 5 in Figure 1, prove l \parallel m from the corresponding angles test.
- In \triangle ABC of Figure 7, \overrightarrow{BD} bisects \angle ABC with D on \overline{AC}, E is on \overline{AB}, and \overline{ED} \parallel \overline{BC}. Prove \angle EBD \cong \angle EDB. What reason takes the place of the converse?
- A transversal t crosses three different lines a, b and c. The angle above each line and right of t is \angle 1 at a, \angle 2 at b and \angle 3 at c. Given \angle 1 \cong \angle 2 and \angle 2 \cong \angle 3, prove a \parallel c, and explain why, unlike Example 6, this needs no parallel postulate.
Reasoning
- To prove l \parallel m from m\angle 1 = 110^\circ and m\angle 7 = 70^\circ, a student writes: “\angle 1 \cong \angle 5 by the Corresponding Angles Theorem, so m\angle 5 = 110^\circ. \angle 5 and \angle 7 form a linear pair, and 110^\circ + 70^\circ = 180^\circ. It all fits, so l \parallel m.” What is wrong? Write a correct argument.
- Which steps of the perpendicular test need l, m and t to lie in one plane?
- Euclid proved the alternate interior test first (Book I, Proposition 27) and the corresponding angles test from it (Proposition 28). This lesson goes the other way. Is either order circular? What must the first test in the chain rest on?
Answers
- The corresponding angles test: both are upper left.
- The alternate exterior angles test: both lie outside the lines, on opposite sides of t.
- The same side interior angles test.
- None. Vertical angles at one crossing are congruent whatever m does.
- Yes, after one chase. \angle 2 and \angle 4 form a linear pair, so m\angle 4 = 180^\circ - m\angle 2 = m\angle 8, and \angle 4, \angle 8 are a congruent corresponding pair.
- None, unless both are right angles. If l \parallel m, then \angle 8 \cong \angle 4 by the Corresponding Angles Theorem, and \angle 4 is a supplement of \angle 2, so \angle 2 and \angle 8 are supplementary. Congruent supplementary angles are 90^\circ each, and then the perpendicular test applies.
- Yes. \angle 8 is vertical to \angle 5, so m\angle 5 = 122^\circ, and 58^\circ + 122^\circ = 180^\circ: \angle 3 and \angle 5 are a supplementary same side interior pair.
- No. They meet on the left of t, at 10^\circ. m\angle 4 = 104^\circ (vertical to \angle 1) and 104^\circ + 86^\circ = 190^\circ, so the test fails. On the left, m\angle 3 = 180^\circ - 104^\circ = 76^\circ and m\angle 5 = 180^\circ - 86^\circ = 94^\circ, adding to 170^\circ. They cannot meet on the right, where a triangle would have two angles adding to 190^\circ; so they meet on the left, in a triangle with third angle 180^\circ - 170^\circ = 10^\circ.
- Parallel at 83^\circ, since 31^\circ + 52^\circ = 83^\circ and Example 6 applies word for word. Not parallel at 85^\circ: if l \parallel m, the bend in Alternate and same side angles, Example 5, would make the angle at X exactly 83^\circ.
- x = 20, and both angles are 86^\circ. \angle 2 and \angle 7 are alternate exterior, so the test wants 5x - 14 = 3x + 26, giving 2x = 40. Check: 5(20) - 14 = 86 = 3(20) + 26.
- x = 15, with m\angle 4 = 109^\circ and m\angle 6 = 71^\circ. A same side interior pair: (7x + 4) + (2x + 41) = 180, so 9x = 135.
- x = 10, with m\angle 3 = 100^\circ and m\angle 5 = 80^\circ. A same side interior pair: x^2 + 6x + 20 = 180, so x^2 + 6x - 160 = 0 and (x - 10)(x + 16) = 0. The root x = -16 gives m\angle 5 = -76^\circ, which is no angle.
- y = 20, with m\angle A = 70^\circ and m\angle B = 110^\circ; the transversal is \overline{AB}. The angles share \overline{AB}, their other sides lie on \overline{AD} and \overline{BC}, and (3y + 10) + (5y + 10) = 180 gives 8y = 160.
- Line (1): \angle 4 \cong \angle 5 (given). (2): \angle 1 \cong \angle 4 (Vertical Angles Theorem: \angle 1 and \angle 4 are opposite at P). (3): \angle 1 \cong \angle 5 (transitive property of congruence: lines 2 and 1). (4): l \parallel m (converse of the Corresponding Angles Theorem: line 3, since \angle 1 and \angle 5 are both upper left).
- Line (1): \overrightarrow{BD} bisects \angle ABC (given). (2): \angle EBD \cong \angle DBC (definition of angle bisector: line 1, with E on \overrightarrow{BA}). (3): \overline{ED} \parallel \overline{BC} (given). (4): \angle EDB \cong \angle DBC (Alternate Interior Angles Theorem: line 3, transversal \overleftrightarrow{BD}). (5): \angle DBC \cong \angle EDB (symmetric property of congruence: line 4). (6): \angle EBD \cong \angle EDB (transitive property of congruence: lines 2 and 5). The forward theorem replaces the converse, because now the lines are given and the angles are wanted. With the converse of the Base Angles Theorem, met later, line 6 then gives EB = ED: the bisector and the parallel make \triangle EBD isosceles (Example 5 of that lesson).
- Line (1): \angle 1 \cong \angle 2 and \angle 2 \cong \angle 3 (given). (2): \angle 1 \cong \angle 3 (transitive property of congruence: line 1). (3): a \parallel c (converse of the Corresponding Angles Theorem: line 2, since \angle 1 and \angle 3 sit in the same position where t crosses a and c). The test needs no postulate, and nothing else was used: the chain joins angles, all on the one transversal t, and never joins parallels. In Example 6, l \parallel n and n \parallel m came along different transversals, so no angle pair on a single line ever linked l to m; only the postulate joins two parallels.
- The first reason is the Corresponding Angles Theorem, whose if-part is l \parallel m: the student assumed the Prove, and “it all fits” shows only that the lines could be parallel. A correct argument: \angle 1 and \angle 3 form a linear pair, so m\angle 3 = 180^\circ - 110^\circ = 70^\circ = m\angle 7, and \angle 3 and \angle 7 are corresponding, so l \parallel m by the converse.
- Two. Calling the right angles corresponding needs a plane, since “the same side of t” means nothing for lines in different planes. And the corresponding angles test ends with “they never meet, so they are parallel”, which is the definition of parallel only for lines in one plane; in space, lines that never meet may be skew.
- Neither is circular, as long as the first test is proved from facts that use no test and no parallel theorem. Euclid proved Proposition 27 from Proposition 16 (an exterior angle of a triangle is larger than either far interior angle). This map proves the corresponding angles test from Proposition 17 in Converses and biconditional definitions, and from Proposition 16 in Indirect proof. After that each test is one vertical angle or linear pair away from the other.