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Geometry · G.CO.C.9

Proving alternate interior angles congruent

Corresponding angles along parallel lines are known to be equal. How do you turn that into a proof that alternate interior angles are equal, one that later proofs can quote?

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Proving lines parallelGeometryAlternate and same side anglesMath 8
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When you reach for it

Twice, in two different ways.

The first time is once only: to earn the theorem. Alternate interior angles states it, shows with a half turn why it holds, and sketches the argument in prose; deductive reasoning (Example 4) deduces it as a chain of reasons, and a flowchart proof (Example 2) draws the same chain as boxes and arrows. Here it is written as a two-column proof, every line resting on a reason already accepted, so that afterwards it can be a reason itself. On this map the order is fixed: the parallel postulate first, then the Corresponding Angles Theorem proved from it, then this theorem proved from that.

The second time is every time after. Whenever a proof has to carry an angle across the gap between two parallel lines, one of its lines will read “Alternate Interior Angles Theorem”, and most of this lesson is about writing that line so it is actually justified.

The procedure

The set-up follows Given, prove and the diagram, with the angles numbered as in Figure 1.

Given. l \parallel m; the line t crosses l and m.

Prove. \angle 3 \cong \angle 6 and \angle 4 \cong \angle 5.

l m t 1 2 3 4 5 6 7 8
Figure 1The diagram for the proof. The arrowheads are the Given; nothing marks the Prove.

The arrowheads on l and m are the Given. No arcs mark \angle 3 and \angle 6, because their congruence is what has to be earned.

Plan. \angle 3 and \angle 6 sit at different crossings, and the only reason on the list that reaches across the gap is the Corresponding Angles Theorem. So go through an angle that is linked to both: \angle 2 is opposite \angle 3 at the same crossing, and it sits in the same position as \angle 6, above its line and to the right of t. For the second pair, \angle 8 does the same job.

StatementsReasons
1. l \parallel mgiven
2. \angle 2 \cong \angle 6Corresponding Angles Theorem: line 1
3. \angle 3 \cong \angle 2Vertical Angles Theorem: \angle 3 and \angle 2 are opposite where t crosses l
4. \angle 3 \cong \angle 6transitive property of congruence: lines 3 and 2
5. \angle 4 \cong \angle 8Corresponding Angles Theorem: line 1
6. \angle 8 \cong \angle 5Vertical Angles Theorem: \angle 8 and \angle 5 are opposite where t crosses m
7. \angle 4 \cong \angle 5transitive property of congruence: lines 5 and 6

Three things about this proof are worth seeing.

The parallel lines are used exactly twice. Lines 2 and 5 need line 1, because the Corresponding Angles Theorem is a conditional whose if-part is l \parallel m. Lines 3 and 6 need no earlier line, only a position read from the diagram: vertical angles are equal at any crossing. So if the lines were not parallel, the proof would break at lines 2 and 5 and nowhere else.

The order inside a line matters. The transitive property joins a \cong b and b \cong c into a \cong c. Line 3 is written \angle 3 \cong \angle 2, not the other way round, so that the shared angle \angle 2 sits in the middle when line 4 joins it to line 2.

It is one route of several. Going through \angle 7 works as well (Exercise 3), and the second pair could come from linear pairs instead. Any route is a proof, provided each reason was proved before this theorem.

Quoting the theorem afterwards. From now on, “Alternate Interior Angles Theorem” is a reason. Like any theorem, it may be cited only when its if-part is already on the page:

  • an earlier line says that two particular lines are parallel, and
  • the two angles are alternate interior angles for those two lines: between them, on opposite sides of a transversal that crosses both, at different crossings. That is a position fact, read from the diagram, as what a diagram may assume allows.

In a figure with several lines, name the transversal in the reason.

Worked examples

Example 1

Alternate exterior angles

Use Figure 1. Given. l \parallel m. Prove. \angle 1 \cong \angle 8.

Solution

Plan. \angle 1 and \angle 8 lie outside the strip between the lines. Each is vertical to an interior angle, \angle 1 to \angle 4 and \angle 8 to \angle 5, and those two are an alternate interior pair. Step in, cross the gap, step out.

StatementsReasons
1. l \parallel mgiven
2. \angle 1 \cong \angle 4Vertical Angles Theorem: \angle 1 and \angle 4 are opposite where t crosses l
3. \angle 4 \cong \angle 5Alternate Interior Angles Theorem: line 1
4. \angle 1 \cong \angle 5transitive property of congruence: lines 2 and 3
5. \angle 5 \cong \angle 8Vertical Angles Theorem: \angle 5 and \angle 8 are opposite where t crosses m
6. \angle 1 \cong \angle 8transitive property of congruence: lines 4 and 5

This is the Alternate Exterior Angles Theorem. That lesson proves it through a corresponding pair instead. Both routes are sound, because both rest in the end on the parallel postulate and neither uses the theorem it is proving.

Example 2

Same side interior angles

Use Figure 1. Given. l \parallel m. Prove. \angle 3 and \angle 5 are supplementary.

Solution

Plan. The Prove is about a sum, so at some point the proof has to turn congruent angles into equal numbers. \angle 5 and \angle 6 form a linear pair, and \angle 6 can be traded for \angle 3.

StatementsReasons
1. l \parallel mgiven
2. \angle 3 \cong \angle 6Alternate Interior Angles Theorem: line 1
3. m\angle 3 = m\angle 6definition of congruent angles: line 2
4. m\angle 6 = m\angle 3symmetric property of equality: line 3
5. m\angle 5 + m\angle 6 = 180^\circLinear Pair Theorem: \angle 5 and \angle 6 form a linear pair along m
6. m\angle 5 + m\angle 3 = 180^\circsubstitution: line 4 into line 5
7. \angle 3 and \angle 5 are supplementarydefinition of supplementary angles: line 6

Line 3 is the hinge: the theorem gives \cong, and only = can be substituted into a sum. With numbers, if m\angle 3 = 71^\circ, then m\angle 6 = 71^\circ and m\angle 5 = 180^\circ - 71^\circ = 109^\circ, and 71 + 109 = 180. This is the Same Side Interior Angles Theorem.

Example 3

Which lines are parallel?

In Figure 2, \overline{AB} \parallel \overline{DC} and m\angle BAC = 38^\circ. A proof contains the line “\angle DAC \cong \angle BCA, Alternate Interior Angles Theorem”. Is the reason right? Then find m\angle DCA with a reason for each step.

38^\circ
 A B C D
Figure 2\overline{AB} \parallel \overline{DC}, and \overline{AD} and \overline{BC} are not parallel.
Solution

Check the if-part. \angle DAC lies between \overline{AD} and \overline{AC}, and \angle BCA between \overline{CB} and \overline{CA}. They are on opposite sides of \overline{AC} and between the lines \overleftrightarrow{AD} and \overleftrightarrow{BC}, so they are alternate interior angles for those two lines. Nothing says \overline{AD} \parallel \overline{BC}, and Figure 2 is drawn with them leaning different ways. The reason fails, and the statement is false in this figure: if the two angles were congruent, the converse of the theorem would make \overline{AD} \parallel \overline{BC}.

The given parallels are \overline{AB} and \overline{DC}, and \overline{AC} crosses both. At A the angle between them is \angle BAC, at C it is \angle DCA, and B and D lie on opposite sides of \overline{AC}.

StatementsReasons
1. \overline{AB} \parallel \overline{DC}given
2. m\angle BAC = 38^\circgiven
3. \angle BAC \cong \angle DCAAlternate Interior Angles Theorem: line 1, transversal \overleftrightarrow{AC}
4. m\angle BAC = m\angle DCAdefinition of congruent angles: line 3
5. 38^\circ = m\angle DCAsubstitution: line 2 into line 4
6. m\angle DCA = 38^\circsymmetric property of equality: line 5

Segments count: the theorem needs the lines through \overline{AB} and \overline{DC} to be parallel, and a diagonal is a transversal. In a parallelogram both pairs of sides are parallel, and then the diagonal gives two alternate interior pairs at once.

Example 4

A path that bounces between two parallels

Given. l \parallel m; A and C are on l; D, B and E are on m, with B between D and E; \angle ABD \cong \angle CBE.

Prove. \angle BAC \cong \angle BCA.

l m A C B D E
Figure 3A path from A to B to C, meeting m at equal angles on the two sides.

Think of a ray of light from A striking a mirror along m at B and going on to C. A mirror sends light off at the same angle it arrived, which is the Given. The Prove says the path meets l at equal angles too.

Solution

Plan. The V is two Z shapes sharing the point B. One Z runs along l from C to A, down \overline{AB}, and along m to D; the other runs along l from A to C, down \overline{CB}, and along m to E. Each Z carries one angle at A or C across to B, and the Given joins them.

StatementsReasons
1. l \parallel mgiven
2. \angle ABD \cong \angle CBEgiven
3. \angle BAC \cong \angle ABDAlternate Interior Angles Theorem: line 1, transversal \overleftrightarrow{AB}
4. \angle BAC \cong \angle CBEtransitive property of congruence: lines 3 and 2
5. \angle CBE \cong \angle BCAAlternate Interior Angles Theorem: line 1, transversal \overleftrightarrow{BC}
6. \angle BAC \cong \angle BCAtransitive property of congruence: lines 4 and 5

Line 3 needs C and D on opposite sides of \overleftrightarrow{AB}, and line 5 needs A and E on opposite sides of \overleftrightarrow{BC}. Both are positions read from Figure 3. Line 5 is written with \angle CBE first so that line 6 can chain it.

Check with numbers. If m\angle ABD = 35^\circ, then m\angle BAC = m\angle BCA = 35^\circ. The three angles at B fill the straight angle along m, so m\angle ABC = 180^\circ - 35^\circ - 35^\circ = 110^\circ.

Example 5

The theorem and its converse in one proof

Use Figure 4. Given. l \parallel m; t crosses l at P and m at Q; \overrightarrow{PR} bisects \angle 3, the angle below l to the left of t; \overrightarrow{QS} bisects \angle 6, the angle above m to the right of t.

Prove. \overleftrightarrow{PR} \parallel \overleftrightarrow{QS}.

l m t P Q R S
Figure 4Each ray bisects one angle of an alternate interior pair. Nothing marks \overrightarrow{PR} and \overrightarrow{QS} as parallel.
Solution

Plan. The Prove is a parallel, so the last line must be a test for parallel lines: the converse of the theorem, proved in Alternate and same side angles and set out with the other tests in proving lines parallel. Its transversal can be t itself. For the lines \overleftrightarrow{PR} and \overleftrightarrow{QS}, the alternate interior pair along t is \angle RPQ and \angle SQP, with R and S on opposite sides of t, and each is half of one of \angle 3 and \angle 6. Halves of equal angles are equal. That a bisector cuts off half its angle is proved once in justifying each step (Exercise 16), so here it is cited in one line.

StatementsReasons
1. l \parallel mgiven
2. \overrightarrow{PR} bisects \angle 3given
3. \overrightarrow{QS} bisects \angle 6given
4. \angle 3 \cong \angle 6Alternate Interior Angles Theorem: line 1
5. m\angle 3 = m\angle 6definition of congruent angles: line 4
6. m\angle RPQ = \tfrac{1}{2}m\angle 3a bisector cuts off half its angle: line 2
7. m\angle SQP = \tfrac{1}{2}m\angle 6a bisector cuts off half its angle: line 3
8. m\angle RPQ = \tfrac{1}{2}m\angle 6substitution: line 5 into line 6
9. \tfrac{1}{2}m\angle 6 = m\angle SQPsymmetric property of equality: line 7
10. m\angle RPQ = m\angle SQPsubstitution: line 9 into line 8
11. \angle RPQ \cong \angle SQPdefinition of congruent angles: line 10
12. \overleftrightarrow{PR} \parallel \overleftrightarrow{QS}converse of the Alternate Interior Angles Theorem: line 11, transversal t

Line 4 uses the theorem, parallel lines to equal angles, and line 12 its converse, equal angles to parallel lines, for a different pair of lines. For instance, if m\angle 3 = 62^\circ, each half is 31^\circ, at P and at Q alike.

Boundaries

Without the parallel lines there is no theorem. Lines 2 and 5 of the proof are the only lines that need l \parallel m, and they are exactly where it fails for lines that are not parallel. Alternate and same side angles shows such a pair measuring 70^\circ and 50^\circ.

The order cannot run backwards. On this map the parallel postulate comes first, then corresponding angles, then this theorem, and Examples 1 and 2 come after it. A proof of the Corresponding Angles Theorem may never cite this theorem, and this theorem may not be proved from the same side interior result of Example 2, which was proved from it. Some courses take a different fact as their postulate; that is allowed, but a proof may not mix two orders.

The pair must belong to the given parallels. Two angles in a Z shape are alternate interior for the two lines that make the Z. If those are not the lines given parallel, the reason fails (Example 3).

The theorem gives \cong, not =. To add, subtract or substitute, go through the definition of congruent angles first, as in Examples 2 and 5.

Where this goes next

The next proof on this branch, proving the triangle angle sum, draws a line through one corner parallel to the opposite side and quotes this theorem twice, the way Example 4 quotes it twice. In a parallelogram a diagonal is a transversal of both pairs of opposite sides, which is how opposite sides are proved congruent and much of Properties of a parallelogram is proved. Run in reverse, as in Example 5, the angle pairs become tests for proving lines parallel. The same reasons laid out as boxes and arrows make a flowchart proof.

Proofs with parallel lines: the chain

This proof is the first of three on this branch, each a two-column proof built on the one before.

  • Across the gap. This lesson: alternate interior angles are equal, from corresponding angles and a pair of vertical angles.
  • Around a triangle. The triangle angle sum adds a line through one vertex, parallel to the opposite side, and lays the three angles along a straight angle, 180^\circ.
  • Outside a triangle. The exterior angle theorem subtracts from that same 180^\circ: an exterior angle equals the sum of the two remote interior angles. A corollary follows: an exterior angle is larger than either remote interior angle.

Each proof cites only what comes before it: the parallel postulate, then the Corresponding Angles Theorem, then alternate interior angles, the angle sum and the exterior angle theorem. The postulate is needed only for uniqueness, at most one parallel through a point off a line, yet everything in the chain rests on it. Euclid’s Elements first uses the postulate in Book I, Proposition 29, which is this lesson’s theorem together with its corresponding and same side partners. The inequality alone can do without: Euclid proved it earlier, as Proposition 16, with no parallel postulate and no angle sum. The converses run the other way, angles in and parallel lines out. They need no postulate either (Propositions 27 and 28), and proving lines parallel proves them that way; alternate interior angles proves them from the postulate instead, which works but uses more than they need.

Once proved, each result is a one-line reason, used by polygon angle sums and by the larger angle opposite the longer side.

Exercises

Reading the proof

Exercises 1 to 3 refer to the proof in The procedure.

  1. Which lines of the proof would still be true if l and m were not parallel? Why?
  2. Why is line 3 written \angle 3 \cong \angle 2 and not \angle 2 \cong \angle 3?
  3. Replace lines 2 to 4 with three lines that go through \angle 7 instead of \angle 2, giving a reason for each.

Naming the reason

In Exercises 4 to 7, a proof about Figure 1 has l \parallel m on line 1. Give the reason for each statement.

  1. \angle 5 \cong \angle 4
  2. \angle 7 \cong \angle 6
  3. m\angle 4 + m\angle 6 = 180^\circ
  4. \angle 2 \cong \angle 7

Which pair of lines?

ABCD is a convex quadrilateral with the diagonal \overline{BD} drawn. For each statement, name the two lines that must be given parallel for the Alternate Interior Angles Theorem to justify it.

  1. \angle ABD \cong \angle CDB
  2. \angle ADB \cong \angle CBD

With numbers

In Exercises 10 and 11, l \parallel m and the angles are numbered as in Figure 1.

  1. m\angle 1 = (3x + 16)^\circ and m\angle 8 = (5x - 20)^\circ. Find x and each angle.
  2. m\angle 3 = (2x + 14)^\circ and m\angle 5 = (4x + 4)^\circ. Find x and each angle.
  3. In Figure 3, m\angle ABD = 28^\circ. Find m\angle BAC, m\angle BCA and m\angle ABC.
  4. In Figure 3, m\angle ABC is four times m\angle ABD. Find m\angle ABD and m\angle BAC.
  5. In a quadrilateral like Figure 2, m\angle BAC = (5y - 7)^\circ and m\angle DCA = (3y + 21)^\circ. Find y and m\angle DCA.
  6. In Figure 4, m\angle 3 = 76^\circ. Find m\angle RPQ and m\angle SQP.

Writing proofs

  1. Use Figure 1. Given. l \parallel m. Prove. \angle 2 \cong \angle 7, using the Alternate Interior Angles Theorem.
  2. Use Figure 1. Given. l \parallel m. Prove. \angle 4 and \angle 6 are supplementary.
  3. Given. In quadrilateral ABCD, \overline{AB} \parallel \overline{DC} and \overline{AD} \parallel \overline{BC}. Prove. \angle DAB \cong \angle BCD.
  4. A zig-zag. Given. l \parallel m; A and P are on l, B and Q are on m, and X is between the lines; P, Q and X are all on the same side of \overleftrightarrow{AB}. Prove. m\angle AXB = m\angle PAX + m\angle QBX. (This is the rule found with numbers in Alternate and same side angles, Example 5.)
  5. In Example 5, would the bisectors of \angle 3 and \angle 5, a same side interior pair, also be parallel? Explain.
Answers
  1. Lines 3 and 6. They cite the Vertical Angles Theorem, which holds at any crossing. Line 1 would be false, lines 2 and 5 cite a theorem whose if-part is line 1, and lines 4 and 7 rest on those.
  2. The transitive property joins a \cong b and b \cong c. With \angle 3 \cong \angle 2 on line 3 and \angle 2 \cong \angle 6 on line 2, the shared \angle 2 is in the middle, so line 4 follows directly. The other order would need an extra line using the symmetric property.
  3. Line (2): \angle 3 \cong \angle 7 (Corresponding Angles Theorem: line 1; both are below their line and left of t). (3): \angle 7 \cong \angle 6 (Vertical Angles Theorem: \angle 7 and \angle 6 are opposite where t crosses m). (4): \angle 3 \cong \angle 6 (transitive property of congruence: lines 2 and 3).
  4. Alternate Interior Angles Theorem: both are between l and m, on opposite sides of t.
  5. Vertical Angles Theorem: they are opposite each other at the crossing with m. No parallel lines are needed.
  6. The Same Side Interior Angles Theorem, proved from this one in Example 2 (there for \angle 3 and \angle 5; the same proof works for \angle 4 and \angle 6).
  7. The Alternate Exterior Angles Theorem, proved in Example 1 for the other pair.
  8. \overleftrightarrow{AB} \parallel \overleftrightarrow{DC}, with transversal \overleftrightarrow{BD}. \angle ABD lies between \overline{BA} and \overline{BD}, \angle CDB between \overline{DC} and \overline{DB}, and A and C are on opposite sides of \overline{BD}.
  9. \overleftrightarrow{AD} \parallel \overleftrightarrow{BC}, with transversal \overleftrightarrow{BD}.
  10. x = 18, and each angle is 70^\circ. \angle 1 \cong \angle 8 (Example 1), so 3x + 16 = 5x - 20 and 2x = 36. Check: 3(18) + 16 = 70 and 5(18) - 20 = 70.
  11. x = 27, m\angle 3 = 68^\circ and m\angle 5 = 112^\circ. The pair is supplementary (Example 2): (2x + 14) + (4x + 4) = 180, so 6x + 18 = 180 and 6x = 162. Check: 68 + 112 = 180.
  12. m\angle BAC = 28^\circ, m\angle BCA = 28^\circ and m\angle ABC = 124^\circ. Line 3 of Example 4 gives \angle BAC \cong \angle ABD; the Given makes m\angle CBE = 28^\circ, and line 5 gives \angle BCA \cong \angle CBE. The three angles at B fill a straight angle, so m\angle ABC = 180^\circ - 28^\circ - 28^\circ = 124^\circ.
  13. m\angle ABD = 30^\circ and m\angle BAC = 30^\circ. Call m\angle ABD = x; then m\angle CBE = x (given) and m\angle ABC = 4x. The angles at B fill a straight angle: x + 4x + x = 180, so 6x = 180 and x = 30. \angle BAC \cong \angle ABD by the theorem.
  14. y = 14 and m\angle DCA = 63^\circ. \angle BAC \cong \angle DCA (Example 3), so 5y - 7 = 3y + 21 and 2y = 28. Check: 5(14) - 7 = 63 and 3(14) + 21 = 63.
  15. m\angle RPQ = 38^\circ and m\angle SQP = 38^\circ. A bisector halves its angle, so m\angle RPQ = 76^\circ \div 2 = 38^\circ. By the theorem m\angle 6 = 76^\circ, and its half is 38^\circ too.
  16. Line (1): l \parallel m (given). (2): \angle 2 \cong \angle 3 (Vertical Angles Theorem: \angle 2 and \angle 3 are opposite where t crosses l). (3): \angle 3 \cong \angle 6 (Alternate Interior Angles Theorem: line 1). (4): \angle 2 \cong \angle 6 (transitive property of congruence: lines 2 and 3). (5): \angle 6 \cong \angle 7 (Vertical Angles Theorem: \angle 6 and \angle 7 are opposite where t crosses m). (6): \angle 2 \cong \angle 7 (transitive property of congruence: lines 4 and 5).
  17. Line (1): l \parallel m (given). (2): \angle 5 \cong \angle 4 (Alternate Interior Angles Theorem: line 1). (3): m\angle 5 = m\angle 4 (definition of congruent angles: line 2). (4): m\angle 5 + m\angle 6 = 180^\circ (Linear Pair Theorem: \angle 5 and \angle 6 form a linear pair along m). (5): m\angle 4 + m\angle 6 = 180^\circ (substitution: line 3 into line 4). (6): \angle 4 and \angle 6 are supplementary (definition of supplementary angles: line 5).
  18. Line (1): \overline{AD} \parallel \overline{BC} (given). (2): \overline{AB} \parallel \overline{DC} (given). (3): \angle DAB and \angle ABC are supplementary (Same Side Interior Angles Theorem, proved in Example 2: line 1, transversal \overleftrightarrow{AB}). (4): \angle ABC and \angle BCD are supplementary (Same Side Interior Angles Theorem: line 2, transversal \overleftrightarrow{BC}). (5): \angle DAB \cong \angle BCD (Congruent Supplements Theorem: lines 3 and 4, both are supplements of \angle ABC).
  19. Draw it with l above m, like the left panel of Figure 7 in Alternate and same side angles. Line (1): l \parallel m (given). (2): n is a line through X parallel to l (a parallel exists through a point off a line: X is not on l). (3): n \parallel m (two lines parallel to the same line are parallel, The parallel postulate Example 4: lines 1 and 2, and n \ne m because X is not on m). Name a point Y on n on the other side of \overleftrightarrow{AX} from P; the figure then shows Y and Q on opposite sides of \overleftrightarrow{BX}, and \overrightarrow{XY} inside \angle AXB. (4): \angle AXY \cong \angle PAX (Alternate Interior Angles Theorem: line 2, transversal \overleftrightarrow{AX}). (5): \angle BXY \cong \angle QBX (Alternate Interior Angles Theorem: line 3, transversal \overleftrightarrow{BX}). (6): m\angle AXY = m\angle PAX (definition of congruent angles: line 4). (7): m\angle BXY = m\angle QBX (definition of congruent angles: line 5). (8): m\angle AXB = m\angle AXY + m\angle BXY (angle addition postulate: \overrightarrow{XY} is inside \angle AXB). (9): m\angle AXB = m\angle PAX + m\angle QBX (substitution: lines 6 and 7 into line 8).
  20. No. \angle 3 and \angle 5 add to 180^\circ (Example 2), so their halves add to 90^\circ. The halves next to t are same side interior angles for the two bisectors with transversal t. If the bisectors were parallel, the Same Side Interior Angles Theorem would make those halves add to 180^\circ, so the bisectors are not parallel. In fact they meet at a right angle, which Same side interior angles proves in Example 6.
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