Geometry · G.CO.C.9

The parallel postulate

A line and a point off it. How many lines through the point never meet the line, and why could nobody prove the answer?

Where you are 4 in Parallel lines and transversals

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

Every proof has to start somewhere. Euclid, around 300 BC, opened his Elements with five postulates: statements accepted without proof, from which everything else is built. Four are short and obvious: you can join two points, extend a line, draw a circle, and all right angles are equal.

The fifth is long, and it is about something nobody can check by drawing: what two lines do far away, off the edge of every page. Yet without it you cannot show that a triangle’s angles add to 180^\circ, that alternate interior angles at parallel lines are equal, or even that two lines parallel to a third are parallel to each other.

For two thousand years mathematicians tried to prove it from the other four, and failed. The failures were the discovery: it cannot be proved, because there are consistent geometries where it is false. It is a choice of plane, and choosing it is what makes school geometry Euclidean geometry.

Definition

Two lines in a plane are parallel when they never meet, however far they are extended. We write m \parallel \ell, and m \nparallel \ell when they do meet.

The parallel postulate. Given a line \ell and a point P not on \ell, there is exactly one line through P parallel to \ell.

Figure 1Through P, exactly one line misses \ell. Every other line through P crosses it.

This is the form most modern books use, often called Playfair’s axiom after John Playfair, who used it in his 1795 edition of Euclid. Euclid’s own wording talks about a transversal, a line crossing two others, and the two interior angles it makes on one side, the ones between the two lines:

Euclid’s form. If a transversal crosses two lines and the two interior angles on one side add to less than 180^\circ, then the two lines, extended far enough, meet on that side.

Figure 2Euclid’s form. The interior angles on the right add to 155^\circ, less than 180^\circ, so the lines meet on the right.

Given the rest of geometry, the two forms say the same thing. Playfair’s is easier to remember; Euclid’s is easier to use with angle measures.

“Exactly one” is two claims: at least one parallel exists, and at most one does. Only the second is the real assumption. The first can be proved without it, as shown below.

Intuition

Stand at P holding a line and turn it slowly. Steep, it crosses \ell close by. As you tip it toward level the crossing slides away: a line tilted 1^\circ from level crosses about 57 times as far away as P is above \ell. At one position the crossing has run off for good. Tip past it and the line crosses \ell again, on the other side.

The postulate says this special position is a single line, not a range of them. Every drawing agrees, but a drawing shows a few centimeters of each line, and the postulate is about what happens infinitely far away. That is why it can be assumed but never measured.

Results

The parallel postulate. Through a point not on a line there is exactly one parallel to it. Equivalently (Euclid’s form), interior angles adding to less than 180^\circ on one side of a transversal force the lines to meet on that side.

These theorems depend on it, and each fails where the postulate fails.

If two parallel lines are cut by a transversal, alternate interior angles are equal, corresponding angles are equal, and same side interior angles add to 180^\circ.
Two lines parallel to the same line are parallel to each other.
A line that crosses one of two parallel lines crosses the other.
The angles of every triangle add to 180^\circ: the Triangle angle sum.
Parallel lines stay the same distance apart all along their length.

The reverse statements, such as “if alternate interior angles are equal, the lines are parallel”, do not need the postulate, as the next section shows.

Why it is true

Strictly, it is assumed, not proved. What can be shown is which half needs assuming, why the two forms agree, and why no proof from the other postulates exists.

At least one parallel needs no postulate. Draw a line through P crossing \ell at Q. At P, build a line m whose alternate interior angle equals the angle at Q (this is how you construct a parallel line). If m met \ell at a point R, triangle PQR would have an exterior angle equal to an interior angle far from it. Euclid’s Proposition 16, proved without the fifth postulate, says an exterior angle is always larger than each far interior angle. So m never meets \ell.

The two forms agree. The parallel m just built has interior angles adding to exactly 180^\circ on each side of PQ. Any other line through P is tilted away from m, so on one side its interior angles add to less than 180^\circ, and Euclid’s form makes it meet \ell there: that is uniqueness. Conversely, a line whose interior angles on one side add to less than 180^\circ is not m, so uniqueness makes it meet \ell, and not on the other side, because two angles of a triangle add to less than 180^\circ (Euclid’s Proposition 17, again proved without the postulate).

What the postulate buys. Let m \parallel \ell through P, with a transversal crossing them at P and Q. Build m' through P with alternate interior angle equal to the one at Q; by the first paragraph, m' \parallel \ell. By the postulate m and m' are one line, so m makes equal alternate interior angles too. That is the alternate interior angles theorem, and every result in the list above is proved by a move like it.

Why no proof exists. Every attempted proof from the other four postulates quietly assumed something equivalent, such as the angle sum of a triangle. In the 1820s and 1830s Nikolai Lobachevsky and János Bolyai, and privately Carl Friedrich Gauss, assumed instead that more than one parallel exists, and never ran into a contradiction. In 1868 Eugenio Beltrami showed why: that geometry has a model built inside ordinary Euclidean geometry.

Figure 3A geometry where the postulate fails. Lines are arcs meeting the rim at right angles, and three of them pass through P without ever reaching \ell.

In one such model the plane is the inside of a circle, the lines are diameters and arcs meeting the rim at right angles, and distance is measured so that the rim is infinitely far away. Euclid’s other postulates hold there, but many lines through P never reach \ell. If the parallel postulate followed from the others it would hold here too.

Worked examples

Example 1

Reading Euclid’s form

A transversal crosses lines m and \ell. On one side of the transversal the two interior angles measure 87^\circ and 91^\circ. Are the lines parallel? If not, on which side do they meet?

Solution

Add the two interior angles on the given side:

87^\circ + 91^\circ = 178^\circ

That is less than 180^\circ, so by Euclid’s form the lines are not parallel, and they meet on this side of the transversal.

Check. Each interior angle on the other side forms a linear pair with a given one, so they measure 180^\circ - 87^\circ = 93^\circ and 180^\circ - 91^\circ = 89^\circ, and 93^\circ + 89^\circ = 182^\circ is more than 180^\circ. The lines do not meet on that side, which agrees: two lines meet at most once.

Example 2

Which tilt is the parallel

A transversal t crosses line \ell at Q and passes through a point P not on \ell. At Q, the interior angle on the right of t is 72^\circ. A line through P makes an interior angle \theta on the right of t. For which \theta is the line parallel to \ell? What happens when \theta = 100^\circ, and when \theta = 115^\circ?

Solution

When the interior angles on the right add to exactly 180^\circ, the alternate interior angles are equal, and the construction in the previous section shows the line never meets \ell:

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

That this line is parallel needs no postulate; that no other \theta works is the postulate.

For \theta = 100^\circ the right-hand sum is 72^\circ + 100^\circ = 172^\circ, less than 180^\circ, so the line meets \ell on the right.

For \theta = 115^\circ the right-hand sum is 187^\circ. On the left the interior angles are 180^\circ - 72^\circ = 108^\circ and 180^\circ - 115^\circ = 65^\circ, which add to 173^\circ, less than 180^\circ, so the line meets \ell on the left.

Example 3

Watching the crossing point run away

Let \ell be the x-axis and P = (0, 3). A line through P with slope s has equation y = sx + 3. Where does it meet \ell when \displaystyle s = -\frac{1}{2}, and when \displaystyle s = -\frac{1}{100}? Which line through P never meets \ell?

Solution

A point is on \ell when its y-coordinate is 0, so set sx + 3 = 0.

For \displaystyle s = -\frac{1}{2}: \displaystyle -\frac{1}{2}x + 3 = 0, so x = 6. The line meets \ell at (6, 0).

For \displaystyle s = -\frac{1}{100}: \displaystyle -\frac{1}{100}x + 3 = 0, so x = 300. The line meets \ell at (300, 0), fifty times further out, although it is barely tilted.

In general \displaystyle x = -\frac{3}{s}, a real number for every s \ne 0, and the vertical line through P meets \ell at the origin. Only s = 0 fails: the line y = 3 would need 3 = 0. So exactly one line through P is parallel to \ell.

This is not a proof: the slope facts behind parallel lines in coordinates rest on similar triangles, which rest on the postulate. It is the Intuition section in numbers.

Example 4

Two lines parallel to a third

Lines \ell and n are different lines, and each is parallel to line m. Show that \ell \parallel n.

Solution

Suppose not. Then \ell and n meet at some point X.

X is on \ell, and \ell never meets m, so X is not on m. Then \ell and n are two lines through X parallel to m. The postulate allows only one, contradicting their being different. So \ell \parallel n.

This is an indirect proof, the natural way to use “only one”: suppose there are two and look for the clash.

Example 5

A line that crosses one parallel crosses the other

Lines \ell and m are parallel. A different line t meets m at a point P. Show that t also meets \ell.

Solution

P is on m, and m never meets \ell, so P is not on \ell. By the postulate, m is the only line through P parallel to \ell. The line t passes through P and is not m, so it is not parallel to \ell: it meets \ell.

So in every transversal problem, the second crossing in the picture really exists.

Example 6

A bent path between parallels

In the left panel of Figure 4, m \parallel \ell. A path runs from A on m to a point X between the lines and on to B on \ell. The angle between m and AX at A is 35^\circ and the angle between \ell and BX at B is 50^\circ. Find x, the angle AXB.

Figure 4A bent path between two parallel lines. The dashed parallel through X, which the postulate guarantees, splits the angle at X in two.

Solution

Nothing connects the angle at X to the lines directly, so add a line: through X, draw the line n parallel to m, a definite line because the postulate says there is exactly one. Since n \parallel m and \ell \parallel m, Example 4 gives n \parallel \ell.

Now AX is a transversal of the parallels m and n. The angle at A and the upper part of the angle at X are alternate interior angles, so the upper part is 35^\circ. In the same way BX is a transversal of n and \ell, and the lower part is 50^\circ. The two parts fill the angle at X:

x = 35^\circ + 50^\circ = 85^\circ

The postulate was used three times: to name n, to know n \parallel \ell, and inside the alternate interior angles theorem.

Boundaries

It is an assumption about the flat plane. In the disc model of Figure 3 there are many parallels through a point, and a triangle’s angles add to less than 180^\circ. On a sphere, with great circles (like the equator) as lines, any two lines meet, so there are no parallels, and a triangle’s angles add to more than 180^\circ. The sphere breaks more than the fifth postulate (its lines are finite loops), which is why the existence proof above fails there.

The point must be off the line. If P is on \ell, every line through P meets \ell at P, so none is parallel to it. (Some books call a line parallel to itself; then \ell is the one parallel, and the count is still one.)

It is a statement about one plane. In space, infinitely many lines through P miss \ell; most are Skew lines, neither parallel nor crossing. The postulate counts only lines in the plane containing \ell and P.

Where this goes next

The postulate turns the angle pairs of parallel lines and transversals from observations into theorems: alternate interior, corresponding and same side interior angles, all proved in proofs with parallel lines. Running those in reverse, which needs no postulate, is proving lines parallel.

Its best-known consequence is the Triangle angle sum: draw the one parallel through a vertex and read the three angles off a straight line (the proof). From there come the angle sums of polygons and much of plane geometry.

And it is the clearest example of what an axiom is: not an obvious truth, but a choice that decides which geometry you are in. How assumptions, definitions and proved results fit together is the subject of definitions, postulates and theorems.

Exercises

Reading the postulate

Line \ell and a point P not on it are given. How many lines through P are parallel to \ell? How many lines through P meet \ell?
Point P lies on line \ell. How many lines through P are parallel to \ell? Explain.
Lines p, q and r all pass through a point X that is not on line \ell, and each of them is parallel to \ell. What can you conclude about p, q and r?

Euclid’s form

A transversal crosses two lines. The two interior angles on one side of the transversal are given. Say whether the lines are parallel, and if not, on which side of the transversal they meet.

88^\circ and 95^\circ
67^\circ and 113^\circ
101^\circ and 76^\circ
The interior angles on one side measure (3x + 10)^\circ and (2x + 20)^\circ, and the lines are parallel. Find x and both angles.
The interior angles on one side measure (4x)^\circ and (5x - 9)^\circ. For which values of x do the lines meet on that side?

Where the lines meet

Line \ell is the x-axis and P = (0, 4). Each line below passes through P.

Where does the line with slope -2 meet \ell?
Where does the line with slope \displaystyle \frac{1}{4} meet \ell?
Where does the line with slope \displaystyle -\frac{1}{1000} meet \ell?
Find the one line through P that never meets \ell, and show that the line with any other slope s meets \ell.
Now let \ell be the line y = 2x + 1 and P = (3, 0). Find the equation of the line through P parallel to \ell.

Chasing angles with the postulate

In Exercises 14 to 16, m \parallel \ell and the path A, X, B bends at X as in Figure 4: the angle at A is between m and AX, the angle at B is between \ell and BX, and x is the angle AXB.

The angle at A is 42^\circ and the angle at B is 63^\circ. Find x.
The angle at A is y^\circ, the angle at B is (2y)^\circ and x = 117^\circ. Find y.
x = 90^\circ and the angle at A is 28^\circ. Find the angle at B.
Lines \ell and n are both parallel to m, and a line t crosses \ell, making an angle of 47^\circ with it. Explain why t must cross n, and find the four angles at that crossing.

Reasoning

Explain why the existence of a parallel through P can be proved without the parallel postulate, while its uniqueness cannot.
Lines \ell and m are parallel, and line t is perpendicular to m. Show that t is perpendicular to \ell.
On a sphere, the “lines” are great circles. Explain why there are no parallel lines on a sphere.
Someone proves the parallel postulate like this: “The angles of any triangle add to 180^\circ, so if two lines through P both missed \ell we could build a triangle whose angles add to more than 180^\circ.” Explain what is wrong with this proof.
Using the parallel postulate, prove that if a transversal makes interior angles on one side adding to less than 180^\circ, the two lines are not parallel.

Answers

Exactly one line through P is parallel to \ell, and infinitely many meet it: every line through P except that one.
None. Every line through P already shares the point P with \ell, so it meets \ell. (If your book counts a line as parallel to itself, the answer is one: \ell itself.)
They are all the same line. The postulate allows only one line through X parallel to \ell.
Not parallel; they meet on the other side. The given side adds to 88^\circ + 95^\circ = 183^\circ, and the other side to 92^\circ + 85^\circ = 177^\circ, which is less than 180^\circ.
Parallel. The interior angles add to 67^\circ + 113^\circ = 180^\circ, so the alternate interior angles are equal, and the lines never meet (no postulate needed for this direction).
Not parallel; they meet on this side, since 101^\circ + 76^\circ = 177^\circ is less than 180^\circ.
x = 30, and the angles are 100^\circ and 80^\circ. Parallel lines need the sum to be 180^\circ: (3x + 10) + (2x + 20) = 180, so 5x = 150.
\displaystyle \frac{9}{5} < x < 21. The lines meet on that side when 4x + (5x - 9) < 180, so 9x < 189 and x < 21. Both angles must be positive, and 5x - 9 > 0 gives \displaystyle x > \frac{9}{5}.
(2, 0). The line is y = -2x + 4, and -2x + 4 = 0 gives x = 2.
(-16, 0). From \displaystyle \frac{1}{4}x + 4 = 0, x = -16.
(4000, 0). From \displaystyle -\frac{1}{1000}x + 4 = 0, x = 4000. A nearly level line still meets \ell, just very far away.
The line y = 4, which would need 4 = 0 to meet \ell. For any slope s \ne 0, sx + 4 = 0 gives \displaystyle x = -\frac{4}{s}, so the line meets \ell at \displaystyle \left(-\frac{4}{s}, 0\right), and the vertical line x = 0 meets it at the origin.
y = 2x - 6. A parallel has the same slope, 2, so y = 2x + b; through (3, 0), 0 = 6 + b and b = -6.
x = 105^\circ. The parallel through X splits x into alternate interior angles of 42^\circ and 63^\circ, and 42 + 63 = 105.
y = 39. As in Example 6, x is the sum of the two angles: y + 2y = 117, so 3y = 117.
62^\circ. The two angles add to x: 28 + b = 90, so b = 62.
47^\circ, 133^\circ, 47^\circ and 133^\circ. By Example 4, \ell \parallel n, so by Example 5 t crosses n. Corresponding angles are equal, so t makes the same angles with n as with \ell: 47^\circ, its supplement 180^\circ - 47^\circ = 133^\circ, and their vertical angles.
A parallel can be built directly: copy the angle at Q to P as an alternate interior angle, and Proposition 16 (an exterior angle of a triangle exceeds each far interior angle, proved without the postulate) shows the new line cannot meet \ell. Uniqueness cannot be proved, because the disc model satisfies all the other postulates and has many parallels.
t meets m, so by Example 5 it also meets \ell and is a transversal of the parallels. Corresponding angles are equal, so t meets \ell at the same 90^\circ as m: t \perp \ell.
A great circle is where the sphere meets a plane through its center. Two such planes meet in a line through the center, which pierces the sphere at two opposite points. Both great circles pass through those points, so any two “lines” meet.
It is circular. The triangle angle sum is itself proved from the parallel postulate, so the argument assumes what it sets out to prove. In the disc model, where the postulate fails, triangles have angle sums below 180^\circ.
Call the lines n and \ell, crossed by the transversal at P on n and Q on \ell. Through P build the line m whose interior angles on that side add to exactly 180^\circ; then m \parallel \ell without any postulate. The angle at P differs, so n \ne m, and by the postulate m is the only parallel to \ell through P. So n \nparallel \ell.