Math 8 · G.CO.C.9 · 8.G.A.5

Alternate and same side angles

A line cuts across two parallel lines. Which angles at one crossing are forced to match an angle at the other, and which are forced to make up 180^\circ with it?

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Corresponding anglesMath 8

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Proving alternate interior angles congruentGeometry

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Geometry11

Points, lines and planes8

Parallel lines and transversals4

Alternate and same side angles

Corresponding angles The parallel postulate Distance from a point to a line

Why this exists

When a single line crosses two others, it makes eight angles, four at each crossing. From vertical angles and linear pairs you already know that the four angles at one crossing come in only two sizes. But nothing so far connects the first crossing to the second. Each crossing is its own little island of information.

Parallel lines build the bridge. If the two lines are parallel, certain angles at the top crossing are forced to equal certain angles at the bottom one, and certain others are forced to add up to a straight angle with them. A single measurement then fixes all eight angles, and the same facts, read backwards, prove that two lines are parallel without ever following them to see whether they meet.

This lesson names those bridges and puts them to work. Alternate interior angles are the most used, because they sit right next to each other in the space between the lines, which is where the triangles and parallelograms of later geometry live. Alternate exterior angles are the same bridge on the outside of the figure. Same side interior angles are the odd one out: they are usually not equal at all, one sharp and one wide, yet they are tied just as tightly, because they always add to 180^\circ.

That last pair is the oldest of them. When Euclid wrote down the assumption that flat geometry rests on, the one now called the parallel postulate, he phrased it with this pair: if the two interior angles on one side add to less than two right angles, the lines meet on that side. So this pair is the one that decides whether two lines are parallel in the first place.

Definition

Two lines l and m are cut by a third line t, the transversal, as in parallel lines and transversals. Number the eight angles as in Figure 1: \angle 1 to \angle 4 at the crossing with l, \angle 5 to \angle 8 at the crossing with m, each time upper left, upper right, lower left, lower right. The arrowheads on l and m are the usual mark for parallel lines.

Figure 1The shaded strip is the interior. Angles 3, 4, 5 and 6 are inside it; angles 1, 2, 7 and 8 are exterior.

Three words describe where an angle sits.

Interior means between the two lines l and m, in the shaded strip: \angle 3, \angle 4, \angle 5 and \angle 6.
Exterior means outside the strip, above l or below m: \angle 1, \angle 2, \angle 7 and \angle 8.
Alternate means on opposite sides of the transversal t; same side means on the same side of it.

Every pair below is made of two angles at different crossings. “Interior” and “exterior” are measured against the two lines; “alternate” and “same side” are measured against the transversal.

Alternate interior angles

Two angles are alternate interior angles when they are both interior and on opposite sides of the transversal. That gives exactly two pairs:

\angle 3 \text{ and } \angle 6 \qquad\qquad \angle 4 \text{ and } \angle 5

There is a quick way to find them by eye. Trace a Z (or a backwards Z) along one line, down the transversal and along the other line. The two angles tucked into the bends of the Z are alternate interior angles.

Figure 2An alternate interior pair sits in the two bends of a Z.

Alternate exterior angles

Two angles are alternate exterior angles when they are both exterior and on opposite sides of the transversal. Again there are exactly two pairs:

\angle 1 \text{ and } \angle 8 \qquad\qquad \angle 2 \text{ and } \angle 7

\angle 1 is above l and to the left of t; \angle 8 is below m and to the right of t. Each word in the name is one condition. Exterior: both are outside the two lines. Alternate: they take turns, one on each side of the transversal.

Figure 3The two alternate exterior pairs, \angle 1 with \angle 8 and \angle 2 with \angle 7. Matching arcs mark equal angles.

Same side interior angles

Two angles are same side interior angles when they are both interior and both on the same side of the transversal:

\angle 3 \text{ and } \angle 5 \qquad\qquad \angle 4 \text{ and } \angle 6

The first pair is on the left of t, the second on the right. Other books call them consecutive interior angles or co-interior angles; the three names mean the same thing.

Figure 4The shaded angles \angle 3 and \angle 5 are one same side interior pair; \angle 4 and \angle 6, on the right of t, are the other.

The last combination, both exterior and on the same side, gives the same side exterior angles \angle 1 and \angle 7, and \angle 2 and \angle 8. They behave like the same side interior pairs and are met again in the examples.

All of these definitions describe position only. They say nothing about size, and they apply whether or not l and m are parallel. Parallel lines are where the pairs become useful.

Intuition

Mark the point O halfway between the two crossings P and Q, and turn the whole picture half a turn, 180^\circ, about O.

Figure 5A half turn about O swaps P and Q, swaps the two lines, and lays each shaded angle on the other.

The turn sends P to Q and Q to P, because O is the midpoint. It sends the transversal onto itself, since the transversal passes through O. And a half turn sends every line to a line parallel to it, so l lands on a line through Q parallel to l. When l \parallel m there is only one such line, m itself. So the turn swaps l and m.

Follow \angle 3 through the turn. It sits at P, below l and to the left of t. After the turn it sits at Q, above m and to the right of t: exactly where \angle 6 is. A turn does not change the size of anything, so the two angles are equal. An alternate interior pair is one angle and its own image under a half turn.

The same turn carries \angle 1, above l on the left, to the place below m on the right, which is \angle 8. That explains both halves of the name “alternate exterior”: the turn keeps an outside angle outside, since the region above l turns into the region below m, but it swaps the two sides of the transversal. Outside stays outside, left becomes right.

Same side interior angles need a different picture. \angle 3 opens downward from l and \angle 5 opens upward from m, so they face each other across the strip. Tilt the transversal so that \angle 3 gets sharper, and \angle 5 gets wider by exactly the same amount, since the transversal turns against both parallel lines at once. Whatever one angle loses, the other gains, so their total never moves. When t is perpendicular to both lines, both angles are right angles and the total is 90^\circ + 90^\circ = 180^\circ. Since the total never moves, it is 180^\circ for every tilt.

Now let the lines stop being parallel. In Figure 6 the interior pair on the right adds to 70^\circ + 80^\circ = 150^\circ, which is 30^\circ short of a straight angle, and the lines lean together on that side until they meet. The triangle they close off has its third angle equal to exactly the 30^\circ that is missing. On the left the pair adds to 210^\circ, and on that side the lines spread apart.

Figure 6When a same side interior pair falls short of 180^\circ, the lines lean together and meet on that side.

Parallel lines are the balanced case: the pair on each side adds to exactly 180^\circ, so the lines lean neither way, and they never meet.

Results

Alternate Interior Angles Theorem. If two parallel lines are cut by a transversal, then the alternate interior angles are congruent.

l \parallel m \quad\Longrightarrow\quad \angle 3 \cong \angle 6 \quad\text{and}\quad \angle 4 \cong \angle 5

Alternate Exterior Angles Theorem. If two parallel lines are cut by a transversal, then the alternate exterior angles are congruent.

l \parallel m \quad\Longrightarrow\quad \angle 1 \cong \angle 8 \quad\text{and}\quad \angle 2 \cong \angle 7

Same Side Interior Angles Theorem. If two parallel lines are cut by a transversal, then each pair of same side interior angles is supplementary.

m\angle 3 + m\angle 5 = 180^\circ

m\angle 4 + m\angle 6 = 180^\circ

The same side exterior pairs are supplementary too: m\angle 1 + m\angle 7 = 180^\circ and m\angle 2 + m\angle 8 = 180^\circ.

The converses. Each theorem also runs the other way. If two lines are cut by a transversal and a pair of alternate interior angles are congruent, or a pair of alternate exterior angles are congruent, or a pair of same side interior angles are supplementary, then the lines are parallel.

\angle 3 \cong \angle 6 \quad\Longrightarrow\quad l \parallel m

The theorems are used to find angles when you know the lines are parallel. The converses are used the other way round, to show that lines are parallel when you know the angles.

Solving problems with parallel lines

Put the pairs together with vertical angles and linear pairs, and the eight angles come in only two sizes. When l \parallel m, any two of the eight angles are either equal or supplementary. When the transversal is not perpendicular you can tell which by looking: two acute angles or two obtuse angles are equal, one of each adds to 180^\circ.

Figure 7Eight angles, only two sizes. One arc marks one size, two arcs the other.

m\angle 2 = m\angle 3 = m\angle 6 = m\angle 7

m\angle 1 = m\angle 4 = m\angle 5 = m\angle 8

m\angle 1 + m\angle 2 = 180^\circ

A real problem rarely hands you the two angles a single fact is about. It gives one angle in one corner and asks about another somewhere else, perhaps as an expression, perhaps at a point that is not on the transversal at all. The way through is to chain pair facts from the angle you know to the one you want, naming the reason for every step:

Pair	In Figure 1	When l \parallel m
corresponding	\angle 1, \angle 5; \angle 2, \angle 6; \angle 3, \angle 7; \angle 4, \angle 8	equal
alternate interior	\angle 3, \angle 6; \angle 4, \angle 5	equal
alternate exterior	\angle 1, \angle 8; \angle 2, \angle 7	equal
same side interior	\angle 3, \angle 5; \angle 4, \angle 6	add to 180^\circ
same side exterior	\angle 1, \angle 7; \angle 2, \angle 8	add to 180^\circ

Find the two parallel lines and the one transversal before anything else, since every fact ties together only angles made by one transversal with two parallels. With expressions, decide equal or supplementary first, and only then write the equation: equal angles get their expressions set equal, supplementary ones get their sum set equal to 180. When the angle you want is at neither crossing, draw a line through its vertex parallel to the given lines; the parallel postulate says there is exactly one, and it makes that vertex a third crossing where the same facts apply.

Why it is true

Alternate interior angles. The argument leans on one fact taken as a starting point about parallel lines: when l \parallel m, corresponding angles are congruent. Angles in the same position at the two crossings, such as \angle 2 and \angle 6, are equal.

Figure 8\angle 3 and \angle 2 are vertical angles; \angle 2 and \angle 6 are corresponding angles.

\angle 3 and \angle 2 sit opposite each other at the same crossing, so they are vertical angles, and

\angle 3 \cong \angle 2

\angle 2 and \angle 6 are corresponding angles, both above their line and to the right of t. Because l \parallel m,

\angle 2 \cong \angle 6

Two angles congruent to the same angle are congruent to each other, so \angle 3 \cong \angle 6.

For the other pair: \angle 4 forms a linear pair with \angle 3 and \angle 5 forms a linear pair with \angle 6, so m\angle 4 = 180^\circ - m\angle 3 and m\angle 5 = 180^\circ - m\angle 6. The right sides are equal, so \angle 4 \cong \angle 5.

Alternate exterior angles. The same two steps in the other order. \angle 1 and \angle 5 are both upper left at their crossings, so they are corresponding angles and \angle 1 \cong \angle 5. \angle 5 and \angle 8 sit opposite each other at the lower crossing, so they are vertical angles and \angle 5 \cong \angle 8. So \angle 1 \cong \angle 8. Going through \angle 6 in the same way gives \angle 2 \cong \angle 6 \cong \angle 7.

Same side interior angles. This time the second step is a linear pair, which always adds to 180^\circ.

Figure 9\angle 5 matches \angle 1, and \angle 1 and \angle 3 fill a straight angle together.

\angle 1 and \angle 5 are corresponding angles, so m\angle 5 = m\angle 1. \angle 1 and \angle 3 sit side by side at the upper crossing and together fill the half of the picture on the left of t, so they form a linear pair:

m\angle 1 + m\angle 3 = 180^\circ

Replace m\angle 1 with the equal m\angle 5:

m\angle 5 + m\angle 3 = 180^\circ

The same argument on the right of t, with \angle 2 and \angle 6 as the corresponding pair, gives m\angle 4 + m\angle 6 = 180^\circ. For the same side exterior pair \angle 1 and \angle 7: m\angle 1 = m\angle 5, and \angle 5 and \angle 7 form a linear pair, so m\angle 1 + m\angle 7 = 180^\circ.

The two sizes. At one crossing, vertical angles are equal and neighbors add to 180^\circ, so the four angles there have at most two sizes, one supplementary to the other. Corresponding angles copy the first crossing onto the second, position for position. So any two of the eight angles are either two of the same size, and equal, or one of each, and supplementary.

The converses. Suppose \angle 3 \cong \angle 6. By the parallel postulate there is exactly one line through P parallel to m; call it l'. By the theorem just proved, l' makes an alternate interior angle with t equal to \angle 6. So l' and l both pass through P and both make the same angle, m\angle 6, with t, on the same side. Only one line through P does that, so l' = l, and l is parallel to m.

The other converses come down to this one. If \angle 1 \cong \angle 8, then since \angle 8 \cong \angle 5 (vertical angles, which needs no parallel lines) and \angle 1 is vertical to \angle 4, we get \angle 4 \cong \angle 5, a congruent alternate interior pair. If m\angle 3 + m\angle 5 = 180^\circ, then since m\angle 5 + m\angle 6 = 180^\circ as well (a linear pair), m\angle 3 = m\angle 6, again a congruent alternate interior pair. Either way l \parallel m.

The forward theorems rest on the parallel postulate. Euclid proves them in Book I, Proposition 29 of the Elements, the first place in the book where he needs the postulate. That is no accident, since the postulate is itself a statement about same side interior angles. The converses do not need it: the proof above borrows the postulate for speed, but Euclid proves them earlier, in Propositions 27 and 28, from the exterior angle inequality alone. Written out as a two-column proof, the alternate interior case is its own skill: proving alternate interior angles congruent.

Worked examples

Example 1

Naming the pairs

Use Figure 1. Name the relationship between each pair of angles.

(a) \angle 4 and \angle 5 \qquad (b) \angle 3 and \angle 5 \qquad (c) \angle 2 and \angle 7 \qquad (d) \angle 3 and \angle 2 \qquad (e) \angle 1 and \angle 7

Solution

Ask two questions of each pair: are both angles between l and m, or both outside, and are they on the same side of t or opposite sides?

(a) \angle 4 is below l and \angle 5 is above m, so both are between the lines. \angle 4 is right of t and \angle 5 is left of it. They are alternate interior angles.

(b) Both are interior, but both are left of t. They are same side interior angles.

(c) \angle 2 is above l and \angle 7 is below m, so both are outside the strip, and they are on opposite sides of t. They are alternate exterior angles.

(d) They are at the same crossing, opposite each other: vertical angles, not one of the two-crossing pairs at all.

(e) Both are exterior and both are left of t. They are same side exterior angles.

Example 2

Finding all eight angles from one

In Figure 10, l \parallel m and m\angle 3 = 62^\circ. Find the other seven angles.

Figure 10One angle is given. Find the other seven.

Solution

Start with the bridge to the other crossing. \angle 3 and \angle 6 are alternate interior angles and the lines are parallel, so

m\angle 6 = 62^\circ

\angle 4 forms a linear pair with \angle 3, so

m\angle 4 = 180^\circ - 62^\circ = 118^\circ

and \angle 5 is the alternate interior partner of \angle 4, so m\angle 5 = 118^\circ.

The rest are vertical angles at their own crossing. \angle 2 is opposite \angle 3, \angle 1 is opposite \angle 4, \angle 7 is opposite \angle 6, and \angle 8 is opposite \angle 5:

m\angle 1 = 118^\circ \qquad m\angle 2 = 62^\circ

m\angle 7 = 62^\circ \qquad m\angle 8 = 118^\circ

Check. The four angles at each crossing go once round a point: 62 + 118 + 62 + 118 = 360. And \angle 3 and \angle 5 are a same side interior pair: 62 + 118 = 180, as it should be.

Example 3

One exterior angle given

In Figure 11, l \parallel m and the angle above l on the left measures 115^\circ. Find x and y.

Figure 11One exterior angle is given. Find x and y.

Solution

The 115^\circ angle is above l and to the left of t. The angle marked x is below m and to the right of t. Both are exterior, at different crossings, on opposite sides of the transversal, so they are alternate exterior angles. The lines are parallel, so the angles are equal:

x = 115^\circ

The angle marked y sits next to x along line m, so the two form a linear pair and add to 180^\circ:

y = 180^\circ - 115^\circ = 65^\circ

Check. y also forms an alternate exterior pair with the angle above l on the right, and that angle makes a linear pair with the 115^\circ angle, so it is 65^\circ as well. Both routes agree.

Example 4

Same side interior angles add, they do not match

(a) In Figure 12, l \parallel m. Find x and y.

Figure 12One interior angle is given. Find x and y.

(b) Two parallel lines are cut by a transversal. A pair of same side interior angles measure (3x + 10)^\circ and (2x + 20)^\circ. Find x and both angles.

Solution

(a) The 108^\circ angle and x both lie between l and m, and both lie on the right of t. They are same side interior angles, and the lines are parallel, so they add to 180^\circ:

x + 108^\circ = 180^\circ \qquad\Longrightarrow\qquad x = 72^\circ

y sits beside x at the lower crossing, on the other side of m. The two form a linear pair, so

y = 180^\circ - 72^\circ = 108^\circ

As a check, y and the 108^\circ angle are both below their lines and on the right of t, so they are corresponding angles and should be equal. They are.

(b) The lines are parallel, so the pair is supplementary. Add the two measures and set the total equal to 180:

\begin{aligned} (3x + 10) + (2x + 20) &= 180 \\ 5x + 30 &= 180 \\ 5x &= 150 \\ x &= 30 \end{aligned}

Substitute back: 3(30) + 10 = 100 and 2(30) + 20 = 80. The angles measure 100^\circ and 80^\circ, and 100 + 80 = 180.

Setting the two expressions equal would give x = 10 and two angles of 40^\circ, which add to 80^\circ, not 180^\circ. The algebra would be fine; the equation would describe a different figure. The relationship decides the equation, so it has to be named before any algebra starts.

Example 5

Alternate interior angles as expressions

In Figure 13, l \parallel m. Find x, and then the measure of every interior angle.

Figure 13An alternate interior pair written as expressions in x.

Solution

The two marked angles are below l and above m, on opposite sides of t: an alternate interior pair, \angle 3 and \angle 6 in the numbering of Figure 1. The lines are parallel, so the two measures are equal:

\begin{aligned} 3x + 14 &= 5x - 22 \\ 36 &= 2x \\ x &= 18 \end{aligned}

Substitute into both expressions, since they must agree:

3(18) + 14 = 68 \qquad\qquad 5(18) - 22 = 68

So \angle 3 and \angle 6 each measure 68^\circ. The other two interior angles need a second relationship. \angle 5 is the same side interior partner of \angle 3, so the two add to 180^\circ and m\angle 5 = 180^\circ - 68^\circ = 112^\circ. \angle 4, the alternate interior partner of \angle 5, measures 112^\circ too. Two relationships in one figure, one making angles equal and one making them add to 180^\circ, and each step names which one it uses.

Example 6

Alternate exterior angles as expressions

In Figure 14, l \parallel m. Find x, then the measure of each of the eight angles.

Figure 14An alternate exterior pair written as expressions in x.

Solution

The upper right angle at l is \angle 2 and the lower left angle at m is \angle 7. Both are exterior and they are on opposite sides of t, so they are alternate exterior angles. The lines are parallel, so set the expressions equal:

\begin{aligned} 3x + 15 &= 5x - 25 \\ 40 &= 2x \\ x &= 20 \end{aligned}

Substitute into both expressions, since they must agree:

3(20) + 15 = 75 \qquad\qquad 5(20) - 25 = 75

Each marked angle measures 75^\circ. At each crossing the angle opposite a marked angle is also 75^\circ, and the two neighbors are 180^\circ - 75^\circ = 105^\circ. So four of the eight angles measure 75^\circ and four measure 105^\circ.

Example 7

Exterior angles on the same side

Lines l \parallel m are numbered as in Figure 1. m\angle 1 = (3x)^\circ and m\angle 7 = (2x + 20)^\circ. Find x and both angles.

Solution

\angle 1 and \angle 7 are both exterior, but they are on the same side of t, so the alternate exterior theorem does not say they are equal. Route through \angle 2 instead.

\angle 7 and \angle 2 are alternate exterior angles, so m\angle 2 = m\angle 7. \angle 1 and \angle 2 form a linear pair along l, so m\angle 1 + m\angle 2 = 180^\circ. Replacing m\angle 2 by m\angle 7:

m\angle 1 + m\angle 7 = 180^\circ

This is the same side exterior fact from the Results, rebuilt from two simpler steps. Now solve:

\begin{aligned} 3x + (2x + 20) &= 180 \\ 5x + 20 &= 180 \\ 5x &= 160 \\ x &= 32 \end{aligned}

So m\angle 1 = 3(32) = 96^\circ and m\angle 7 = 2(32) + 20 = 84^\circ, and 96 + 84 = 180.

Example 8

Are the lines parallel?

Two lines l and m are cut by a transversal, with the angles numbered as in Figure 1. The lines are not known to be parallel. Decide whether l \parallel m.

(a) m\angle 3 = 71^\circ and m\angle 6 = 73^\circ

(b) m\angle 4 = 109^\circ and m\angle 6 = 71^\circ

(c) m\angle 3 = 83^\circ and m\angle 5 = 99^\circ. If the lines are not parallel, on which side of the transversal do they meet?

(d) In Figure 15, the angle above l on the left measures 110^\circ and the angle below m on the right measures 108^\circ.

Figure 15Two lines that are nearly parallel. The angles decide whether they are.

Solution

(a) \angle 3 and \angle 6 are alternate interior angles, and they are not equal. If the lines were parallel, the theorem would force them to be equal. So the lines are not parallel, however close to parallel they look. Two degrees is too little to see in a drawing, and that is the point: the numbers decide, not the eye.

(b) These two are not an alternate interior pair: \angle 4 and \angle 6 are both right of t. Move one of them first. \angle 3 forms a linear pair with \angle 4, so

m\angle 3 = 180^\circ - 109^\circ = 71^\circ

Now \angle 3 and \angle 6 are an alternate interior pair and both measure 71^\circ. By the converse, l \parallel m. (Equally, \angle 4 and \angle 6 are a same side interior pair, and 109 + 71 = 180.)

(c) \angle 3 and \angle 5 are a same side interior pair, and by the converse the lines are parallel exactly when the pair adds to 180^\circ. Here

83^\circ + 99^\circ = 182^\circ

which is not 180^\circ, so the lines are not parallel. On the other side of the transversal each interior angle forms a linear pair with one of these, so the interior angles there are 180^\circ - 83^\circ = 97^\circ and 180^\circ - 99^\circ = 81^\circ, adding to 178^\circ. That is the side where the pair falls short of 180^\circ, so the lines lean together and meet there, on the right of t, making a triangle whose third angle is the 2^\circ shortfall.

(d) The two marked angles are \angle 1 and \angle 8: exterior, at different crossings, on opposite sides of t, so an alternate exterior pair. If l were parallel to m, the theorem would force them to be equal. They are not, since 110^\circ \ne 108^\circ, so

l \nparallel m

Look closely and you can see m tilt a little in Figure 15, by exactly the 2^\circ between the two measures, and the gap between the lines closes up to the left, where they meet far off the page. With measures of 110^\circ and 109.9^\circ no drawing would show the tilt, but the test would give the same answer.

Example 9

The value that makes the lines parallel

A transversal cuts lines l and m, and a pair of alternate exterior angles measure (7x - 8)^\circ and (4x + 19)^\circ. For what value of x is l \parallel m?

Solution

Here the lines are not given as parallel; the question is when they are. By the converse, the lines are parallel exactly when these alternate exterior angles are equal:

\begin{aligned} 7x - 8 &= 4x + 19 \\ 3x &= 27 \\ x &= 9 \end{aligned}

Check. 7(9) - 8 = 55 and 4(9) + 19 = 55, so both angles measure 55^\circ and the lines are parallel. For any other value of x the two measures differ, and the lines are not parallel.

Example 10

Proving the converse in one case

Lines l and m are cut by a transversal t, numbered as in Figure 1, and \angle 2 \cong \angle 7. Prove that l \parallel m.

Solution

The idea is to trade \angle 7 for an angle that sits in the same position as \angle 2, so that a fact about corresponding angles applies.

\angle 7 and \angle 6 are vertical angles at the crossing with m, so \angle 7 \cong \angle 6. This holds for any two crossing lines, parallel or not, which is why it may be used before anything is known about l and m.

We are given \angle 2 \cong \angle 7, so \angle 2 \cong \angle 6, since both are congruent to \angle 7.

\angle 2 and \angle 6 are both upper right at their crossings, so they are corresponding angles. When a transversal makes corresponding angles congruent, the lines it cuts are parallel. Therefore l \parallel m.

This is how a converse gets used in a formal proof: one congruence you are given, one you get for free from vertical angles, and a named reason to finish.

Example 11

A bend between two parallel lines

In the left panel of Figure 16, l \parallel m. Segments from A on l and B on m meet at a point X between the lines. Find x, the angle at X.

Figure 16A bend between two parallel lines. The dashed helper line n on the right turns it into two Z shapes.

Solution

No two of the marked angles form an alternate interior pair: the figure has no single transversal that crosses both lines at A and B. So add a line. Through X, draw n parallel to l, as in the right panel. Since l \parallel m, n is parallel to m as well.

Now \overline{AX} is a transversal of l and n. The 40^\circ angle at A and the angle between \overline{XA} and n are on opposite sides of \overline{AX} and between l and n, so they are alternate interior angles. The angle at X above n is 40^\circ.

In the same way \overline{BX} is a transversal of n and m, and the angle at X below n is the alternate interior partner of the 35^\circ angle at B. It is 35^\circ.

The helper line splits the angle at X into these two parts, so

x = 40^\circ + 35^\circ = 75^\circ

The same argument works for any bend: the angle at the bend is the sum of the two angles the segments make with the parallel lines.

Example 12

Angles of a trapezoid

In trapezoid ABCD of Figure 17, \overline{AB} \parallel \overline{DC}, m\angle D = 64^\circ and m\angle C = 71^\circ. Find m\angle A and m\angle B.

Figure 17In a trapezoid each leg is a transversal of the two parallel sides.

Solution

Read the leg \overline{AD} as a transversal of the two parallel sides. \angle A and \angle D are both between \overline{AB} and \overline{DC} and both on the same side of \overline{AD}, so they are same side interior angles, and supplementary:

m\angle A = 180^\circ - 64^\circ = 116^\circ

The other leg, \overline{BC}, is a transversal as well, with \angle B and \angle C as its same side interior pair:

m\angle B = 180^\circ - 71^\circ = 109^\circ

Check. The four angles of a quadrilateral add to 360^\circ: 116 + 109 + 71 + 64 = 360. They do.

Notice that \angle A and \angle B are not a same side interior pair, since the side joining them, \overline{AB}, is not a transversal of two parallel lines here. Nothing forces them to add to 180^\circ, and they do not: 116 + 109 = 225. In a parallelogram, where both pairs of sides are parallel, every two neighboring angles are a same side interior pair.

Example 13

The bisectors meet at a right angle

Parallel lines are cut by a transversal at P and Q. The two interior angles on one side of the transversal are each cut in half by a ray, an angle bisector, and the two rays meet at R, as in Figure 18. Show that m\angle PRQ = 90^\circ, whatever the angle of the transversal.

Figure 18The bisectors of a same side interior pair always meet at a right angle.

Solution

Call the two interior angles \alpha at P and \beta at Q. They are a same side interior pair of parallel lines, so

\alpha + \beta = 180^\circ

The bisectors split them in half, so in triangle PQR the angle at P is \tfrac{\alpha}{2} and the angle at Q is \tfrac{\beta}{2}. Their sum is

\frac{\alpha}{2} + \frac{\beta}{2} = \frac{\alpha + \beta}{2} = \frac{180^\circ}{2} = 90^\circ

The three angles of a triangle add to 180^\circ, by the triangle angle sum (proved in the next example), so the third angle is

m\angle PRQ = 180^\circ - 90^\circ = 90^\circ

Nothing in the argument used a particular value of \alpha, so it holds for every transversal. In Figure 18, \alpha = 70^\circ and \beta = 110^\circ, giving halves of 35^\circ and 55^\circ.

Example 14

Why the angles of a triangle add to 180^\circ

Use alternate interior angles to show that the three angles of any triangle add to 180^\circ.

Solution

Take any triangle ABC, and call its angles a at A, b at B and c at C. Let m be the line through B and C, and draw the line l through A parallel to m, as in Figure 19. The parallel postulate says there is exactly one such line.

Figure 19The triangle’s angles at B and C reappear at A, beside its angle there.

\overline{AB} is a transversal of l and m. The triangle’s angle at B and the angle between l and \overline{AB} at A, on the left, lie between the lines and on opposite sides of \overline{AB}. They are alternate interior angles, so the angle at A on the left is b.

\overline{AC} is also a transversal. For the same reason, the angle between \overline{AC} and l at A, on the right, equals c.

At A the three angles b, a and c now sit side by side along the straight line l, filling it. A straight angle measures 180^\circ, so

a + b + c = 180^\circ

Nothing about the triangle was special, so this holds for every triangle. This is the triangle angle sum, and alternate interior angles are the whole proof. In geometries where the parallel postulate fails, triangle angles do not add to 180^\circ.

Boundaries

The lines must be parallel, and you must know it. Every pair is defined for any two lines, but the equalities and the 180^\circ sums hold only when the lines are parallel. In Figure 20 the lines tilt toward each other, and the alternate interior pair measure 70^\circ and 50^\circ. They differ by exactly the angle at which l and m meet, far off the page (Exercise 34). Without l \parallel m, vertical angles and linear pairs still hold at each crossing separately, but nothing links one crossing to the other. A figure that merely looks parallel is not enough; the arrowheads or the words are what you are allowed to use.

Figure 20When l \nparallel m, the pair need not match. Here they are 70^\circ and 50^\circ.

Name the transversal first. “Alternate”, “same side”, “interior” and “exterior” are measured against one particular transversal and one particular pair of lines. In a figure with many lines, such as a parallelogram with its diagonal drawn, the same angle can belong to several pairs, and two angles that look like a Z may be formed by lines that are not the parallel ones. With two transversals crossing the same pair of lines, an angle made by one has no partner made by the other. Say which two lines are parallel and which line crosses them before claiming anything.

Exterior means outside the two lines, not outside the transversal. The transversal only decides alternate or same side. \angle 1 and \angle 7 are both exterior but on the same side, and they are supplementary, not equal.

Equal or supplementary depends only on position. Which pair two angles form comes from where they sit, never from their measures, and the equation follows from the pair. The wrong equation still produces a tidy answer, as Example 4 showed.

A perpendicular transversal is the one case you cannot sort by eye. Then all eight angles are 90^\circ, equal and supplementary at once, which is the perpendicular transversal theorem. It is also the one case where a same side interior pair is equal.

The converse needs an exact fit and one true pair. Angles that are close to equal, or add to nearly 180^\circ, prove nothing. And the two angles must be a pair for the transversal that crosses l and m; two equal angles anywhere else in the figure prove nothing about l and m.

It is a fact about the flat plane. The proofs rest on the parallel postulate, through the corresponding angles fact and through the uniqueness of the parallel line. On a sphere, where the “lines” are great circles, there are no parallel lines at all, and two great circles that both cross the equator at right angles, so that a same side interior pair adds to exactly 180^\circ, still meet at the poles.

Where this goes next

These pairs are the tool that carries an angle across the gap between two parallel lines, and most of what follows is that one move used cleverly. Example 14 is already the triangle angle sum, and from it comes the exterior angle theorem. In a parallelogram each side and each diagonal is a transversal of a pair of opposite sides, so neighboring angles are supplementary and the angles across a diagonal match, which is how the properties of a parallelogram are proved.

Run the other way, the converses are the standard tests for proving lines parallel. Written out with a reason on every line, a chain like Example 2 is already a proof, and the formal version, with each step justified in a two-column proof, starts from proving alternate interior angles congruent. Together with corresponding angles, these facts let you find every angle in a figure full of parallel lines; the same habit of letting one angle decide the rest, in figures with no parallel lines at all, is the work of finding unknown angle measures.

Exercises

Naming the pairs

Use Figure 1 for Exercises 1 to 6.

Name the angle that forms an alternate interior pair with \angle 4.
Name the angle that forms an alternate exterior pair with \angle 2.
Name the two pairs of same side interior angles.
Which of the eight angles are exterior angles, and which are interior?
Name the relationship between each pair: alternate interior, alternate exterior, same side interior, same side exterior, corresponding, or vertical. (a) \angle 1 and \angle 8 (b) \angle 3 and \angle 6 (c) \angle 2 and \angle 6 (d) \angle 1 and \angle 7 (e) \angle 4 and \angle 6
In Figure 1, l \parallel m. List every angle that is supplementary to \angle 6.

Finding measures

In Exercises 7 to 12, l \parallel m and the angles are numbered as in Figure 1.

m\angle 3 = 47^\circ. Find m\angle 6 and m\angle 5.
m\angle 8 = 131^\circ. Find m\angle 1 and m\angle 2.
m\angle 7 = 22^\circ. Find all eight angles.
m\angle 4 = 90^\circ. Find all eight angles.
m\angle 6 is twice m\angle 5. Find m\angle 3 and m\angle 4.
The two angles of a same side interior pair differ by 44^\circ. Find both angles.

Using algebra

In Exercises 13 to 22, the two lines are parallel, and where angles are numbered they are numbered as in Figure 1.

m\angle 3 = (4x + 9)^\circ and m\angle 6 = (6x - 21)^\circ. Find x and the measure of each angle.
A pair of alternate exterior angles measure (2x + 40)^\circ and (5x - 20)^\circ. Find x and the measure of each angle.
A pair of same side interior angles measure (5x + 12)^\circ and (3x + 8)^\circ. Find x and both angles.
m\angle 3 = (2y)^\circ and m\angle 5 = (3y + 30)^\circ. Find y, m\angle 3 and m\angle 5.
m\angle 3 = (6x + 4)^\circ and m\angle 6 = (4x + 30)^\circ. Find x and m\angle 5.
m\angle 2 = (3x + 5)^\circ and m\angle 8 = (4x + 7)^\circ. Find x, m\angle 2 and m\angle 8.
m\angle 2 = (3x)^\circ, m\angle 6 = (2x + 24)^\circ and m\angle 8 = (5y - 2)^\circ. Find x and y.
A pair of same side interior angles measure (x^2)^\circ and (3x + 50)^\circ. Find every value of x that works, and the two angles each one gives.
m\angle 3 = (x^2 + 4x)^\circ and m\angle 6 = (10x + 16)^\circ. Find every value of x that gives a real figure, and the angle it gives.
m\angle 4 = (x^2)^\circ and m\angle 5 = (4x + 60)^\circ. Find every value of x that solves the equation and the angle each one gives. Which one fits Figure 1, where \angle 4 is obtuse?

Are the lines parallel?

In Exercises 23 to 28, two lines are cut by a transversal, and where angles are numbered they are numbered as in Figure 1, but the lines are not known to be parallel.

m\angle 3 = 64^\circ and m\angle 5 = 118^\circ. Is l \parallel m? Why?
m\angle 2 = 58^\circ and m\angle 8 = 122^\circ. Is l \parallel m? Why?
A pair of alternate exterior angles measure 97^\circ and 83^\circ. Are the lines parallel?
On one side of the transversal the interior angles measure 88^\circ and 93^\circ. Are the lines parallel? If not, on which side of the transversal do they meet?
m\angle 4 = (5x - 8)^\circ and m\angle 5 = (3x + 30)^\circ. What value of x makes l \parallel m, and what is each angle then?
A pair of same side interior angles measure (2x + 15)^\circ and (3x - 10)^\circ. What value of x makes the lines parallel, and what is each angle then?

In other figures

In a figure like Figure 16, l \parallel m, the angle at A is 28^\circ and the angle at B is 51^\circ. Find the angle at X.
In a figure like Figure 16, l \parallel m, the angle at X is a right angle and the angle at A is 34^\circ. Find the angle at B.
ABCD is a parallelogram, so \overline{AB} \parallel \overline{DC} and \overline{AD} \parallel \overline{BC}, and the diagonal \overline{AC} is drawn. m\angle BAC = 42^\circ and m\angle BCA = 67^\circ. Find m\angle DCA, m\angle DAC and m\angle B.
Two parallel lines cross two other parallel lines, and the four crossings are the corners of a parallelogram. One angle of the parallelogram is 65^\circ. Find the other three.
In trapezoid ABCD with \overline{AB} \parallel \overline{DC}, m\angle A = (2y)^\circ and m\angle D = (y + 30)^\circ. Find y and both angles.
In Figure 20, m\angle 3 = 70^\circ and m\angle 6 = 50^\circ, and the lines l and m meet at a point R off to the right. Find the angle at R between l and m.
Two lines that are not parallel are cut by a transversal. On one side the interior angles are 65^\circ and 85^\circ, and the lines meet on that side. Find the angle at which they meet.
In triangle ABC, a line is drawn through A parallel to \overline{BC}. It makes an angle of 48^\circ with \overline{AB} and an angle of 57^\circ with \overline{AC}, on either side of the triangle’s angle at A, as in Figure 19. Find the three angles of the triangle.

Reasoning

A transversal is perpendicular to one of two parallel lines. Explain why it is perpendicular to the other as well.
The proof in “Why it is true” shows \angle 3 \cong \angle 6 in two steps. Which step fails when l and m are not parallel, and why?
The lines are parallel. Can both angles of a same side interior pair be acute? Explain.
Two parallel lines are cut by a transversal. Explain why the two angles in an alternate exterior pair are either both acute, both right, or both obtuse.
Prove the Same Side Interior Angles Theorem again, this time using alternate interior angles instead of corresponding angles.
In Figure 1, l \parallel m. Explain why any two of the eight angles are either equal or supplementary.
Two lines are cut by a transversal and \angle 3 \cong \angle 6. Show that the other pair of alternate interior angles, \angle 4 and \angle 5, are congruent too, without assuming the lines are parallel.
In the right panel of Figure 16, explain why the helper line n had to be drawn parallel to l. What goes wrong with the argument if it is just any line through X?
Two lines are cut by a transversal, and a pair of same side interior angles are both right angles. Must the lines be parallel? Explain.
Lines l and m are not parallel. Explain why a transversal cannot make a pair of alternate exterior angles congruent.

Answers

\angle 5. Both are between the lines, and they are on opposite sides of t.
\angle 7. It is exterior, at the other crossing, and on the opposite side of t from \angle 2.
\angle 3 and \angle 5, and \angle 4 and \angle 6. Each pair lies between l and m on one side of t.
Exterior: \angle 1, \angle 2, \angle 7, \angle 8, which lie above l or below m. Interior: \angle 3, \angle 4, \angle 5, \angle 6, which lie between the lines.
(a) Alternate exterior. (b) Alternate interior: both are between the lines, on opposite sides of t. (c) Corresponding: both are upper right at their crossings. (d) Same side exterior: both are outside the lines, on the left of t. (e) Same side interior: both are between the lines, on the right of t.
\angle 1, \angle 4, \angle 5 and \angle 8. \angle 6 is one of the acute angles, and each obtuse angle adds to 180^\circ with it: \angle 5 and \angle 8 by linear pairs at the lower crossing, \angle 4 as its same side interior partner, and \angle 1 because it equals \angle 5 as a corresponding angle.
m\angle 6 = 47^\circ and m\angle 5 = 133^\circ. \angle 6 is the alternate interior partner of \angle 3; \angle 5 is the same side interior partner of \angle 3, so it is 180^\circ - 47^\circ.
m\angle 1 = 131^\circ and m\angle 2 = 49^\circ. \angle 1 and \angle 8 are alternate exterior angles; \angle 2 forms a linear pair with \angle 1, so it is 180^\circ - 131^\circ.
m\angle 2 = m\angle 3 = m\angle 6 = m\angle 7 = 22^\circ and m\angle 1 = m\angle 4 = m\angle 5 = m\angle 8 = 158^\circ. \angle 7 is one of the acute angles, so the other three acute ones match it, and each obtuse one is 180^\circ - 22^\circ.
All eight are 90^\circ. The neighbors of \angle 4 are 180^\circ - 90^\circ = 90^\circ, so all four at the upper crossing are right angles, and \angle 5, the alternate interior partner of \angle 4, is 90^\circ, so the same holds at the lower crossing. This is the perpendicular case, where the two sizes become one.
m\angle 3 = 120^\circ and m\angle 4 = 60^\circ. \angle 5 and \angle 6 form a linear pair, so x + 2x = 180 with x = m\angle 5, giving m\angle 5 = 60^\circ and m\angle 6 = 120^\circ. Then \angle 3 matches \angle 6 and \angle 4 matches \angle 5, as alternate interior pairs.
68^\circ and 112^\circ. The pair adds to 180^\circ, so with x the smaller angle, x + (x + 44) = 180, 2x = 136 and x = 68. Check: 112 - 68 = 44.
x = 15, and each angle is 69^\circ. Alternate interior angles are equal: 4x + 9 = 6x - 21, so 2x = 30. Check: 4(15) + 9 = 69 and 6(15) - 21 = 69.
x = 20, and each angle is 80^\circ. Alternate exterior angles are equal: 2x + 40 = 5x - 20, so 60 = 3x. Check: 2(20) + 40 = 80 and 5(20) - 20 = 80.
x = 20, and the angles are 112^\circ and 68^\circ. The pair is supplementary: (5x + 12) + (3x + 8) = 180, so 8x + 20 = 180 and 8x = 160. Then 5(20) + 12 = 112 and 3(20) + 8 = 68.
y = 30, m\angle 3 = 60^\circ and m\angle 5 = 120^\circ. \angle 3 and \angle 5 are a same side interior pair, so 2y + (3y + 30) = 180, giving 5y = 150.
x = 13 and m\angle 5 = 98^\circ. \angle 3 and \angle 6 are alternate interior, so equal: 6x + 4 = 4x + 30 gives 2x = 26, and m\angle 3 = 6(13) + 4 = 82^\circ. \angle 5 is the same side interior partner of \angle 3, so m\angle 5 = 180^\circ - 82^\circ = 98^\circ.
x = 24, m\angle 2 = 77^\circ and m\angle 8 = 103^\circ. \angle 2 and \angle 8 are both exterior and both right of t, a same side exterior pair, so (3x + 5) + (4x + 7) = 180 and 7x = 168.
x = 24 and y = 22. \angle 2 and \angle 6 are corresponding: 3x = 2x + 24, so m\angle 6 = 72^\circ. \angle 6 and \angle 8 form a linear pair, so 5y - 2 = 180 - 72 = 108 and 5y = 110.
x = 10, giving 100^\circ and 80^\circ, or x = -13, giving 169^\circ and 11^\circ. The pair adds to 180: x^2 + 3x + 50 = 180, so x^2 + 3x - 130 = 0 and (x - 10)(x + 13) = 0. Both values give two positive angles, 10^2 = 100 with 3(10) + 50 = 80, and (-13)^2 = 169 with 3(-13) + 50 = 11.
x = 8, giving 96^\circ. Setting the alternate interior pair equal, x^2 + 4x = 10x + 16, so x^2 - 6x - 16 = 0 and (x - 8)(x + 2) = 0. The root x = -2 gives (-2)^2 + 4(-2) = -4, a negative angle, so it is rejected. Check: 8^2 + 4(8) = 96 and 10(8) + 16 = 96.
x = 10, giving 100^\circ, or x = -6, giving 36^\circ. \angle 4 and \angle 5 are alternate interior angles, so x^2 = 4x + 60, x^2 - 4x - 60 = 0 and (x - 10)(x + 6) = 0. Both give a positive angle, but only x = 10 fits Figure 1: there \angle 4 is obtuse, and 36^\circ is not.
No. \angle 3 and \angle 5 are a same side interior pair, and 64 + 118 = 182, not 180, so by the theorem the lines cannot be parallel.
Yes. \angle 1 forms a linear pair with \angle 2, so m\angle 1 = 180^\circ - 58^\circ = 122^\circ. Then \angle 1 and \angle 8 are alternate exterior angles of equal measure, so by the converse l \parallel m.
No. If the lines were parallel these alternate exterior angles would be equal, and 97^\circ \ne 83^\circ. The fact that 97 + 83 = 180 does not help: adding to 180^\circ is the rule for same side pairs, not alternate ones.
No; they meet on the other side of the transversal. 88^\circ + 93^\circ = 181^\circ, not 180^\circ. On the other side the interior angles are 92^\circ and 87^\circ, adding to 179^\circ, and the lines lean together on the side where the pair falls short.
x = 19, and each angle is 87^\circ. \angle 4 and \angle 5 are alternate interior, so the lines are parallel exactly when 5x - 8 = 3x + 30, which gives 2x = 38. Check: 5(19) - 8 = 87 and 3(19) + 30 = 87.
x = 35, and the angles are 85^\circ and 95^\circ. The lines are parallel exactly when (2x + 15) + (3x - 10) = 180, so 5x + 5 = 180 and 5x = 175.
79^\circ. Draw the helper line through X parallel to l, as in Example 11; the angle at X splits into 28^\circ and 51^\circ, and 28^\circ + 51^\circ = 79^\circ.
56^\circ. As in Example 11, the angle at X is the sum of the angles at A and B, so the angle at B is 90^\circ - 34^\circ = 56^\circ.
m\angle DCA = 42^\circ, m\angle DAC = 67^\circ, m\angle B = 71^\circ. For \overline{AB} \parallel \overline{DC} the diagonal is a transversal and \angle BAC, \angle DCA are alternate interior. For \overline{AD} \parallel \overline{BC}, \angle DAC and \angle BCA are. In triangle ABC, m\angle B = 180^\circ - 42^\circ - 67^\circ.
115^\circ, 65^\circ and 115^\circ, going round. Each side of the parallelogram is a transversal of the other pair of sides, so neighboring angles are same side interior angles and add to 180^\circ: 180^\circ - 65^\circ = 115^\circ. The angle opposite the 65^\circ one is next to a 115^\circ angle, so it is 180^\circ - 115^\circ = 65^\circ.
y = 50, so m\angle A = 100^\circ and m\angle D = 80^\circ. Leg \overline{AD} is a transversal of the parallel sides, so \angle A and \angle D are same side interior: 2y + (y + 30) = 180, giving 3y = 150.
20^\circ. Triangle PQR has angle \angle 4 = 180^\circ - 70^\circ = 110^\circ at P and \angle 6 = 50^\circ at Q, so the angle at R is 180^\circ - 110^\circ - 50^\circ = 20^\circ, which is 70^\circ - 50^\circ, the amount by which the pair fail to match.
30^\circ. The two interior angles and the meeting angle are the three angles of a triangle, so the meeting angle is 180^\circ - 65^\circ - 85^\circ = 30^\circ, the amount by which the pair falls short of 180^\circ.
m\angle B = 48^\circ, m\angle C = 57^\circ and m\angle A = 75^\circ. Alternate interior angles give B and C, and the three angles at A fill a straight line: 180^\circ - 48^\circ - 57^\circ = 75^\circ.
The transversal makes a 90^\circ angle with the first line, and that angle has an alternate interior partner at the second line. The lines are parallel, so the partner is also 90^\circ, and the transversal is perpendicular to the second line.
The second step, \angle 2 \cong \angle 6. Corresponding angles are congruent only when the lines are parallel. The first step, \angle 3 \cong \angle 2, uses vertical angles at a single crossing and holds for any two lines.
No. Two acute angles are each less than 90^\circ, so together they make less than 180^\circ, but a same side interior pair of parallel lines adds to exactly 180^\circ. At most one of the pair is acute, and if one is acute the other is obtuse.
At parallel lines alternate exterior angles are congruent, so they have the same measure. Two angles of the same measure are both less than, both equal to, or both greater than 90^\circ.
\angle 4 and \angle 5 are alternate interior angles, so m\angle 4 = m\angle 5. \angle 3 and \angle 4 sit side by side along l and form a linear pair, so m\angle 3 + m\angle 4 = 180^\circ. Replacing m\angle 4 with m\angle 5 gives m\angle 3 + m\angle 5 = 180^\circ.
At one crossing, vertical angles are equal and neighbors add to 180^\circ, so the four angles there have at most two sizes, one supplementary to the other. The lines are parallel, so corresponding angles are equal and the other crossing has the same two sizes in the same places. Any two of the eight are then two of the same size (equal) or one of each (supplementary).
\angle 4 forms a linear pair with \angle 3 and \angle 5 forms a linear pair with \angle 6, so m\angle 4 = 180^\circ - m\angle 3 and m\angle 5 = 180^\circ - m\angle 6. Since m\angle 3 = m\angle 6, the right sides are equal, so \angle 4 \cong \angle 5. Only linear pairs were used, so parallel lines were never needed.
The two parts at X are found as alternate interior angles, and that fact needs n to be parallel to l (for the upper part) and to m (for the lower part). With any other line through X the two parts still add to the angle at X, but nothing ties either one to the angles at A and B.
Yes. The two angles add to 90^\circ + 90^\circ = 180^\circ, and supplementary same side interior angles force the lines to be parallel, by the converse. Both lines are perpendicular to the transversal. This is a fact about the flat plane: on a sphere, as the Boundaries note, two such great circles still meet.
If a transversal made a pair of alternate exterior angles congruent, the converse would force l \parallel m, and the lines are given as not parallel. So no such pair can be congruent.