3. Parallel lines and the angle sum¶

Every schoolchild learns that the angles of a triangle add up to \(180°\). Why should that be? It is not true of a triangle drawn on a ball, so it cannot be a fact about "all triangles everywhere": it must come from something special about the flat plane. This chapter finds that something. It studies parallel lines and the angles a crossing line makes with them, proves the angle sum and its consequences for every polygon, uses those to show that only three regular polygons can tile a floor, and then looks hard at the one assumption all of it rests on: Euclid's fifth postulate, in the simpler form that John Playfair made standard.

In this chapter

You will learn:

the angles a transversal makes with two lines (corresponding, alternate, co-interior), and which facts about them need the parallel postulate and which do not
parallelograms, rectangles and the fact that parallel lines stay the same distance apart
the proof that the angles of a triangle add to \(180°\), the exterior angle theorem, and where the parallel postulate is hiding in the proof
angle sums of polygons, convex or not, the \(360°\) rule for exterior angles, and star polygons
why only triangles, squares and hexagons tile the plane as regular tilings; semi-regular tilings; which other shapes tile
Playfair's axiom, its equivalence with Euclid's fifth postulate, and a list of other statements equivalent to it

You need: Chapter 1 and Chapter 2, especially the exterior angle inequality (Theorem 2.2). For a gentler version, see the Topics lessons Parallel and perpendicular lines, Angles with parallel lines, Angles in a triangle add to 180°, Angles inside and outside any polygon and Tiling the floor.

Levels: 3.1 to 3.3 are beginner to intermediate; 3.4 is intermediate; 3.5 is intermediate to advanced.

3.1 Parallel lines and transversals¶

The eight angles¶

Two lines \(\ell\) and \(m\) in a plane are parallel if they never meet. A third line \(t\) that crosses both is a transversal. At each crossing it makes four angles, eight in all.

The pairs have names, which describe where they sit:

corresponding angles are in the same position at the two crossings: \(a\) and \(e\), \(b\) and \(f\), \(c\) and \(g\), \(d\) and \(h\) (an "F" shape);
alternate (interior) angles are between the two lines, on opposite sides of the transversal: \(c\) and \(e\), \(d\) and \(f\) (a "Z" shape);
co-interior angles (also called same-side interior or consecutive interior) are between the lines on the same side of the transversal: \(c\) and \(f\), \(d\) and \(e\) (a "C" or "U" shape).

At each crossing, vertical angles are equal and linear pairs add to \(180°\) (Chapter 1), so \(a = c\), \(b = d\), \(a + b = 180°\), and the same at the second crossing. That means that as soon as one of these is true:

\[ \text{corresponding angles equal} \iff \text{alternate angles equal} \iff \text{co-interior angles add to } 180°, \]

all three are true. (For instance, if alternate angles \(c = e\), then corresponding \(a = c = e\), and co-interior \(c + f = e + f = 180°\).) So we can state everything in terms of alternate angles.

Equal alternate angles force parallel lines¶

Theorem 3.1 (Euclid I.27). If a transversal makes equal alternate angles with two lines, the lines are parallel.

Proof. Suppose the transversal crosses \(\ell\) at \(P\) and \(m\) at \(Q\), the alternate angles are equal, and yet the lines meet, at a point \(R\). Then \(P\), \(Q\), \(R\) form a triangle. One of the two alternate angles, say the one at \(Q\), is an exterior angle of this triangle (it forms a linear pair with the interior angle \(\angle PQR\)), and the other, at \(P\), is an interior angle not next to it. By the exterior angle inequality (Theorem 2.2) the exterior angle is strictly larger. But they are equal. Contradiction; so the lines do not meet. \(\blacksquare\)

The figure is drawn the only way it can be: with the angles visibly unequal. A correct picture of the impossible situation cannot be drawn, which is what the proof says.

This theorem uses nothing about parallels: it holds in neutral geometry. It gives, at once:

Corollary 3.2. (a) Two lines perpendicular to the same line are parallel. (b) Through a point \(P\) not on a line \(\ell\) there is at least one line parallel to \(\ell\).

Proof. (a) A transversal perpendicular to both makes alternate angles of \(90°\). (b) Drop the perpendicular \(PQ\) from \(P\) to \(\ell\) (Chapter 2), then draw the line \(m\) through \(P\) perpendicular to \(PQ\). By (a), \(m \parallel \ell\). \(\blacksquare\)

So parallels always exist. The question that Euclid's fifth postulate settles is whether there is only one.

The parallel postulate¶

We will use Euclid's fifth postulate in the form that John Playfair put in his 1795 textbook Elements of Geometry (the idea is older; Proclus states something close to it in the 5th century):

Playfair's axiom. Through a point not on a line there is at most one line parallel to it.

Together with Corollary 3.2(b): exactly one. Section 3.5 proves that Playfair's axiom and Euclid's fifth postulate say the same thing. From now on, every theorem that uses Playfair's axiom will say so, by the words "(uses the parallel postulate)".

Theorem 3.3 (Euclid I.29; uses the parallel postulate). If two lines are parallel, then a transversal makes equal alternate angles with them, equal corresponding angles, and co-interior angles adding to \(180°\).

Proof. Let \(\ell \parallel m\), and let the transversal meet \(\ell\) at \(P\) and \(m\) at \(Q\). Through \(P\) draw the line \(\ell'\) that makes, with the transversal, an alternate angle equal to the one \(m\) makes at \(Q\) (copy the angle, Chapter 2). By Theorem 3.1, \(\ell' \parallel m\). Now \(\ell\) and \(\ell'\) are both lines through \(P\) parallel to \(m\). By Playfair's axiom they are the same line. So \(\ell\) itself makes equal alternate angles with \(m\). The other two statements follow as shown above. \(\blacksquare\)

Theorems 3.1 and 3.3 are converses of each other. Together: two lines cut by a transversal are parallel if and only if the alternate angles are equal. One direction is free, the other costs the parallel postulate.

Theorem 3.4 (uses the parallel postulate). Two lines parallel to the same line are parallel to each other.

Proof. Let \(a \parallel c\) and \(b \parallel c\), with \(a \ne b\). If \(a\) and \(b\) met at a point \(P\), they would be two different lines through \(P\) parallel to \(c\), against Playfair. \(\blacksquare\)

Angle chasing¶

With these facts you can find unknown angles in a figure by following equal angles around it, a skill called angle chasing.

Worked example. A transversal crosses two parallel lines. One alternate angle is \((3x + 12)°\) and the other is \((5x - 30)°\). Find all eight angles. Alternate angles are equal: \(3x + 12 = 5x - 30\), so \(x = 21\) and the angle is \(75°\). Four of the eight angles are \(75°\) and the other four are \(180° - 75° = 105°\).

Worked example (the bent path). Two parallel lines, and a path from \(A\) on one to \(B\) on the other with one bend at \(X\) between them. The path makes angle \(\alpha\) with the first line at \(A\) and \(\beta\) with the second at \(B\) (both measured on the same side). What is the angle at the bend?

Draw through \(X\) a third line parallel to both (it exists by Corollary 3.2, and by Theorem 3.4 it is parallel to the second line too). It splits \(\angle AXB\) into two parts. The upper part is alternate to \(\alpha\) (the lines through \(A\) and \(X\) are parallel, transversal \(AX\)), so it equals \(\alpha\); the lower part equals \(\beta\) in the same way. So

\[ \angle AXB = \alpha + \beta . \]

With \(\alpha = 34°\) and \(\beta = 40°\), the bend is \(74°\). (In the figure the angles are about \(34°\) and \(40°\) and the bend measures about \(74°\).)

Parallelograms¶

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. Chapter 2 promised its properties.

Theorem 3.5 (uses the parallel postulate). In a parallelogram \(ABCD\):

opposite sides are equal: \(AB = CD\), \(BC = DA\);
opposite angles are equal, and consecutive angles add to \(180°\);
the diagonals bisect each other.

Proof. Draw diagonal \(AC\). Because \(AB \parallel DC\), the alternate angles \(\angle BAC\) and \(\angle DCA\) are equal (Theorem 3.3, transversal \(AC\)). Because \(AD \parallel BC\), the alternate angles \(\angle DAC\) and \(\angle BCA\) are equal. Triangles \(ABC\) and \(CDA\) share \(AC\), so they are congruent by ASA. That gives \(AB = CD\), \(BC = DA\) and \(\angle B = \angle D\). Using diagonal \(BD\) instead gives \(\angle A = \angle C\). Consecutive angles such as \(\angle A\) and \(\angle B\) are co-interior for the parallels \(AD\), \(BC\), so they add to \(180°\). For (3), let the diagonals meet at \(M\). Triangles \(ABM\) and \(CDM\) have \(AB = CD\) and two pairs of equal alternate angles (\(\angle BAM = \angle DCM\), \(\angle ABM = \angle CDM\)), so they are congruent by ASA, and \(AM = CM\), \(BM = DM\). \(\blacksquare\)

The converses are useful tests.

Theorem 3.6. A quadrilateral is a parallelogram if any one of these holds:

both pairs of opposite sides are equal;
one pair of opposite sides is both equal and parallel;
the diagonals bisect each other.

Proof. (1) Triangles \(ABC\) and \(CDA\) are congruent by SSS, so \(\angle BAC = \angle DCA\): equal alternate angles, so \(AB \parallel DC\) by Theorem 3.1. Similarly \(AD \parallel BC\). (3) By Theorem 2.17, opposite sides are equal; apply (1). (2) If \(AB = CD\) and \(AB \parallel CD\), then \(\angle BAC = \angle DCA\) (Theorem 3.3), so triangles \(BAC\) and \(DCA\) are congruent by SAS, giving \(\angle BCA = \angle DAC\), which are alternate angles for \(AD\) and \(BC\); so \(AD \parallel BC\). \(\blacksquare\)

Notice that tests (1) and (3) used only Theorem 3.1, so they hold in neutral geometry; the properties in Theorem 3.5 do not.

Rectangles. A rectangle has four right angles. Its opposite sides are perpendicular to the same line, so they are parallel (Corollary 3.2): every rectangle is a parallelogram, with opposite sides equal and diagonals bisecting each other. Its diagonals are also equal: triangles \(ABC\) and \(BAD\) have \(AB\) shared, right angles at \(B\) and \(A\), and \(BC = AD\), so they are congruent by SAS, and \(AC = BD\). Conversely (using the parallel postulate), a parallelogram with one right angle has four, since consecutive angles add to \(180°\).

Parallel lines are everywhere the same distance apart¶

The distance from a point to a line is the length of the perpendicular from the point (the shortest segment, Corollary 2.12).

Theorem 3.7 (uses the parallel postulate). If \(\ell \parallel m\), then all points of \(m\) are the same distance from \(\ell\).

Proof. Take two points \(P\), \(Q\) on \(m\), with feet of perpendiculars \(P'\), \(Q'\) on \(\ell\). The lines \(PP'\) and \(QQ'\) are both perpendicular to \(\ell\), so they are parallel (Corollary 3.2). So \(PQQ'P'\) has both pairs of opposite sides parallel: it is a parallelogram, and by Theorem 3.5, \(PP' = QQ'\). \(\blacksquare\)

This is why railway tracks and the lines on a page can be parallel and stay a constant width apart, and it is what Chapter 4 uses when it says that triangles "between the same parallels" have the same height. On a sphere the corresponding statement fails: the circles of latitude stay a constant distance from the equator, but they are not "straight" on the sphere (Chapter 16).

3.2 The angle sum is 180°¶

The proof¶

Theorem 3.8 (Euclid I.32; uses the parallel postulate). The angles of any triangle add up to \(180°\).

Proof. Let the triangle be \(ABC\) with angles \(\alpha\), \(\beta\), \(\gamma\). Through \(C\) draw the line parallel to \(AB\) (it exists, and by Playfair's axiom it is the only one). The transversal \(CA\) makes alternate angles with the two parallels: the angle between the new line and \(CA\), on the left of \(C\), equals \(\alpha\). In the same way, the angle on the right of \(C\), between the new line and \(CB\), equals \(\beta\) (transversal \(CB\)). The three angles at \(C\), namely \(\alpha\), \(\gamma\), \(\beta\), sit side by side along the straight line, so

\[ \alpha + \beta + \gamma = 180° . \qquad \blacksquare \]

Proclus, quoting the history of geometry by Eudemus (4th century BC), credits this proof to the Pythagoreans; it is essentially the proof in Euclid. The figure in the proof was drawn with a computer, and the three angles came out as \(36.38°\), \(49.40°\) and \(94.22°\), adding to \(180.00°\). That is a check, not a proof; the proof covers every triangle.

Consequences¶

Theorem 3.9 (exterior angle theorem; uses the parallel postulate). An exterior angle of a triangle equals the sum of the two interior angles not next to it.

Proof. The exterior angle at \(C\) and \(\gamma\) form a linear pair, so the exterior angle is \(180° - \gamma\), which by Theorem 3.8 is \(\alpha + \beta\). \(\blacksquare\)

This upgrades Theorem 2.2 ("larger than each") to an equation.

Corollaries.

Each angle of an equilateral triangle is \(60°\) (three equal angles, Corollary 2.7, adding to \(180°\)).
The two acute angles of a right triangle add to \(90°\): they are complementary.
The base angles of an isosceles triangle with apex angle \(\theta\) are each \((180° - \theta)/2 = 90° - \theta/2\). An isosceles right triangle has angles \(45°, 45°, 90°\).
If two angles of one triangle equal two angles of another, the third angles are equal too. (So AAS follows from ASA at once. Chapter 2 proved AAS without this, the hard way, because it is also true in neutral geometry.)
The angles of a quadrilateral add to \(360°\) (cut it into two triangles by a diagonal; see 3.3).

Worked example. The angles of a triangle are in the ratio \(2 : 3 : 7\). Then \(12\) parts make \(180°\), a part is \(15°\), and the angles are \(30°, 45°, 105°\). Check: \(30 + 45 + 105 = 180\).

Worked example. In triangle \(ABC\), the exterior angle at \(C\) is \(128°\) and \(\angle A = 47°\). Then \(\angle B = 128° - 47° = 81°\) and \(\angle C = 180° - 128° = 52°\). Check: \(47 + 81 + 52 = 180\).

Worked example. An isosceles triangle has one angle of \(40°\). What are the others? It depends which angle is \(40°\). If it is the apex, the base angles are \((180 - 40)/2 = 70°\) each. If it is a base angle, the other base angle is \(40°\) too and the apex is \(100°\). Two answers: \(40, 70, 70\) or \(40, 40, 100\).

Where the parallel postulate hides¶

The proof used the parallel postulate once, through Theorem 3.3, when it said that the line through \(C\) makes alternate angles equal to \(\alpha\) and \(\beta\). Without the postulate we could still draw a parallel through \(C\) (Corollary 3.2), but not know that it makes those angles.

Is the postulate really needed, or just convenient? Look at a sphere, where the "straight lines" are great circles (like the equator and the meridians, the shortest routes on the surface).

Start at the north pole, walk south along a meridian to the equator, turn left through \(90°\), walk a quarter of the way around the equator, turn left through \(90°\) again and walk north back to the pole. Your path is a triangle made of three great-circle arcs, and each of its angles is \(90°\): the angle sum is \(270°\). On a sphere there are no parallel lines at all (any two great circles meet), so the parallel postulate fails, and with it the angle sum.

The sphere breaks some of the other axioms too (two points such as the poles lie on many great circles), so it does not yet prove that the parallel postulate is independent of the rest. That needs the hyperbolic plane of Chapter 17, where every axiom of neutral geometry holds, the parallel postulate fails, and every triangle has angle sum less than \(180°\). In Chapter 16 you will meet Girard's theorem: on a sphere of radius \(R\), a triangle of area \(A\) has angle sum \(180°\) plus \(A/R^2\) radians. For the triangle above, \(A\) is one eighth of the sphere, \(4\pi R^2/8\), and \(A/R^2 = \pi/2\) radians \(= 90°\), giving \(270°\).

Small triangles on a big sphere are very nearly flat. A triangle of area \(1000\ \text{km}^2\) on the Earth (radius \(6371\) km) has angle sum exceeding \(180°\) by \(1000/6371^2 \approx 2.46 \times 10^{-5}\) radians, about \(5''\) (seconds of arc); the same triangle on Mars (radius \(3389.5\) km) exceeds it by about \(18''\). Land surveyors who map large regions do correct for this. Carl Friedrich Gauss's survey of the kingdom of Hanover in the 1820s included a large triangle of three mountain peaks (Brocken, Hoher Hagen and Inselsberg). It is often said he was testing whether space itself is Euclidean; historians disagree about whether that was his aim. Either way, the angles agreed with Euclid within the accuracy of the instruments.

3.3 Polygons' angle sums¶

Polygons¶

A polygon is a closed chain of segments \(P_1P_2, P_2P_3, \ldots, P_nP_1\) (its sides) that do not cross one another, and no two neighbouring sides on one line. The \(P_i\) are its vertices; with \(n\) sides it is an \(n\)-gon (triangle, quadrilateral, pentagon, hexagon, heptagon, octagon, nonagon, decagon, ..., dodecagon for 12). A diagonal joins two vertices that are not neighbours. A polygon is convex if every diagonal lies inside it; otherwise concave. A polygon is regular if all its sides are equal and all its angles are equal.

How many diagonals? Each vertex can be joined to \(n - 3\) others (not itself and not its two neighbours), giving \(n(n-3)\) ends, and each diagonal has two ends, so there are \(n(n-3)/2\) diagonals. A hexagon has \(6 \cdot 3/2 = 9\), a 20-gon has \(20 \cdot 17/2 = 170\).

Interior angles¶

Theorem 3.10 (uses the parallel postulate). The interior angles of any \(n\)-gon add up to \((n - 2) \times 180°\).

Proof for convex polygons. From one vertex \(P_1\) draw the \(n - 3\) diagonals to the vertices that are not its neighbours. They cut the polygon into \(n - 2\) triangles (\(P_1P_2P_3\), \(P_1P_3P_4\), ..., \(P_1P_{n-1}P_n\)). The angles of these triangles exactly make up the angles of the polygon: at \(P_1\) the triangles' angles fit together to make the polygon's angle, and at every other vertex one or two triangles meet and fill the polygon's angle. So the polygon's angle sum is \((n - 2) \times 180°\). \(\blacksquare\)

Proof for every polygon. Level: intermediate. A concave polygon may have no vertex from which all diagonals lie inside (try the right-hand figure). Instead we use:

Lemma 3.11. Every polygon with at least four sides has a diagonal lying entirely inside it.

Proof. Let \(V\) be the lowest vertex (if several, the leftmost of those), with neighbours \(U\) and \(W\). The interior angle at \(V\) is less than \(180°\), because the whole polygon lies on or above the horizontal line through \(V\). If segment \(UW\) lies inside the polygon, it is a diagonal. If not, some vertices of the polygon lie inside triangle \(UVW\) (or on \(UW\)); among them take the one, \(X\), farthest from the line \(UW\). Then the segment \(VX\) meets no side (a side crossing it would have to have an end even farther from \(UW\) inside the triangle), so \(VX\) is a diagonal. \(\blacksquare\)

Now induction on \(n\) (a proof that a statement holds for \(n = 3\) and that if it holds for all smaller polygons it holds for \(n\); see the Algebra book's Chapter 8). For \(n = 3\) it is Theorem 3.8. For \(n \ge 4\), a diagonal from Lemma 3.11 cuts the \(n\)-gon into a \(k\)-gon and an \((n - k + 2)\)-gon, for some \(k\) between 3 and \(n - 1\). Their angle sums are \((k - 2) \cdot 180°\) and \((n - k) \cdot 180°\), and together they make up the angles of the original polygon, so its sum is \((k - 2 + n - k) \cdot 180° = (n - 2) \cdot 180°\). \(\blacksquare\)

The same induction shows that every polygon can be triangulated (cut into triangles by diagonals) and that every triangulation of an \(n\)-gon has exactly \(n - 2\) triangles. The right-hand figure, a concave 7-gon, has 5.

polygon	\(n\)	angle sum	each angle if regular	each exterior angle if regular
triangle	3	\(180°\)	\(60°\)	\(120°\)
quadrilateral	4	\(360°\)	\(90°\)	\(90°\)
pentagon	5	\(540°\)	\(108°\)	\(72°\)
hexagon	6	\(720°\)	\(120°\)	\(60°\)
octagon	8	\(1080°\)	\(135°\)	\(45°\)
decagon	10	\(1440°\)	\(144°\)	\(36°\)
dodecagon	12	\(1800°\)	\(150°\)	\(30°\)

A regular \(n\)-gon has each angle equal to \(\dfrac{(n-2) \cdot 180°}{n} = 180° - \dfrac{360°}{n}\).

Exterior angles¶

At each vertex of a convex polygon, extend one side; the angle between the extension and the next side is the exterior angle there, and it is \(180°\) minus the interior angle.

Theorem 3.12 (uses the parallel postulate). The exterior angles of a convex polygon, one at each vertex, add up to \(360°\).

Proof. There are \(n\) of them, so their sum is \(n \cdot 180° - (n - 2) \cdot 180° = 360°\). \(\blacksquare\)

There is a more vivid way to see it. Walk around the polygon. At each corner you turn through the exterior angle. When you are back where you started you face the way you started, having turned once around: \(360°\). That argument is not yet a proof (what does "turning once around" mean, exactly?), but it generalises in a way the algebra does not. For a non-convex polygon, count the turns at the dents as negative (you turn the other way); the total is still \(360°\). For a curve, the "total turning" is the integral of curvature, and that it equals \(360°\) for a simple closed curve in the plane is Hopf's Umlaufsatz (1935), a seed of the Gauss-Bonnet theorem of Chapter 19.

Worked example. Each interior angle of a regular polygon is \(156°\). How many sides? Each exterior angle is \(180° - 156° = 24°\), and they add to \(360°\), so \(n = 360/24 = 15\). Check: \((15 - 2) \cdot 180 / 15 = 156\).

Worked example. Is there a regular polygon with interior angles of \(130°\)? The exterior angle would be \(50°\), and \(360/50 = 7.2\) is not a whole number. No.

Worked example. A pentagon has angles \(x\), \(x + 10°\), \(x + 20°\), \(x + 30°\), \(x + 40°\). The sum is \(540°\), so \(5x + 100 = 540\), \(x = 88°\), and the angles are \(88°, 98°, 108°, 118°, 128°\).

Stars¶

Join every second vertex of five points around a circle and you get a five-pointed star, a pentagram. Here is a pretty fact that holds for every such star, regular or not.

Theorem 3.13 (uses the parallel postulate). The five tip angles of a five-pointed star add up to \(180°\).

Proof. Label the tips \(A, B, C, D, E\) in order around the star, so the five segments are \(AC\), \(CE\), \(EB\), \(BD\), \(DA\). The segment \(EB\) crosses \(AC\) at a point \(P\) and \(AD\) at a point \(Q\), cutting off the triangle \(APQ\) at the tip \(A\). Its angles add to \(180°\): \(\angle A + \angle APQ + \angle AQP = 180°\). Now \(\angle APQ\) is an exterior angle of triangle \(PCE\) (the side \(CP\) extended beyond \(P\) runs to \(A\)), so by Theorem 3.9 it equals \(\angle C + \angle E\). Likewise \(\angle AQP\) is an exterior angle of triangle \(QDB\), so it equals \(\angle B + \angle D\). Substituting, \(\angle A + \angle B + \angle C + \angle D + \angle E = 180°\). \(\blacksquare\)

For the regular pentagram each tip is \(36°\). Walking around the star you turn \(144°\) at each tip, \(720°\) in total: you go around the centre twice. The number of times is the star's turning number, and for any closed polygonal path the sum of the turning angles is \(360°\) times it.

3.4 Tilings of the plane by regular polygons¶

Level: intermediate.

A tiling (or tessellation) covers the whole plane with copies of shapes, with no gaps and no overlaps. Floors, walls, honeycombs and the paving stones of old cities are pieces of tilings, and the question of which shapes can do it turns out to be deep.

Only three regular tilings¶

A tiling is regular if it uses copies of one regular polygon, meeting edge to edge (each edge of each tile is a whole edge of exactly one neighbour).

Theorem 3.14. There are exactly three regular tilings: by equilateral triangles, by squares, and by regular hexagons.

Proof. Suppose regular \(n\)-gons tile edge to edge and \(k\) of them meet at each vertex. Each angle is \(180° - 360°/n\), and the angles around a vertex add to \(360°\) (Theorem 1.4), so

\[ k\left(180° - \frac{360°}{n}\right) = 360° . \]

Divide by \(180°\) and multiply by \(n\): \(k(n - 2) = 2n\), so \(kn - 2k - 2n = 0\), and adding 4 to both sides,

\[ (k - 2)(n - 2) = 4 . \]

Here \(n \ge 3\) and \(k \ge 3\) (two polygons cannot fill \(360°\), since each angle is less than \(180°\)), so both factors are positive whole numbers with product 4: \(1 \times 4\), \(2 \times 2\) or \(4 \times 1\). That gives \((k, n) = (3, 6)\), \((4, 4)\), \((6, 3)\): three hexagons, four squares or six triangles at each vertex. Each really does tile, as the figure shows. \(\blacksquare\)

The factoring trick, \(kn - 2k - 2n + 4 = (k-2)(n-2)\), turns a question about angles into a question about the divisors of 4. You will see it again for the Platonic solids in Chapter 11, where the equation becomes \((k - 2)(n - 2) < 4\).

Why is the failure of pentagons so visible? Three regular pentagons around a point use \(3 \times 108° = 324°\) and leave a \(36°\) gap that nothing regular fits, while four would need \(432°\).

What if tiles need not meet edge to edge (like bricks in a wall)? A vertex of one tile might then lie in the middle of another tile's edge, where the angles must add to \(180°\) instead of \(360°\). Then the angle \(180° - 360°/n\) must divide \(360°\) or \(180°\), and for \(n = 5\) (\(108°\)) or \(n \ge 7\) (between \(128°\) and \(180°\)) it divides neither. So the same three polygons are the only regular ones that tile at all; only the arrangements change.

Semi-regular tilings¶

Allow several kinds of regular polygon, still edge to edge, and ask that every vertex look the same (the same polygons in the same cyclic order). These are the semi-regular or Archimedean tilings. If polygons with \(n_1, \ldots, n_k\) sides meet at a vertex, then

\[ \sum_{i=1}^{k} \left(180° - \frac{360°}{n_i}\right) = 360°, \quad\text{that is}\quad \sum_{i=1}^{k} \left(\frac12 - \frac{1}{n_i}\right) = 1 . \]

A short computer search (or a patient afternoon) finds that this equation has exactly 17 solutions in whole numbers \(n_i \ge 3\), which can be arranged around a vertex in 21 different ways. For example \(4.8.8\) (a square and two octagons: \(90 + 135 + 135 = 360\)), \(3.12.12\), \(4.6.12\), \(3.3.3.3.6\), and the curious \(3.7.42\) (\(60° + 128\tfrac47° + 171\tfrac37° = 360°\)). But a vertex that works is not yet a tiling: the neighbouring vertices must work too, and most of the 21 cannot be continued. Exactly eight can, giving the eight semi-regular tilings: \(3.3.3.3.6\), \(3.3.3.4.4\), \(3.3.4.3.4\), \(3.4.6.4\), \(3.6.3.6\), \(3.12.12\), \(4.6.12\), \(4.8.8\). Johannes Kepler listed them, with the three regular ones, in Harmonices Mundi (1619).

Exercise 11 shows why \(5.5.10\) (two pentagons and a decagon: \(108 + 108 + 144 = 360\)) fits around one vertex but can never be continued into a tiling.

Other shapes that tile¶

Drop regularity. Which single shapes tile the plane?

Every triangle. Two copies of any triangle, one turned half a turn, make a parallelogram, and parallelograms tile in rows.
Every quadrilateral, even a concave one. Turn copies half a turn about the midpoints of the sides. At each corner of the tiling the four different angles of the quadrilateral meet once each, and they add to \(360°\) (Theorem 3.10), so they fit exactly. The copies repeat by sliding along the two diagonals of the original.

Pentagons: not all. Convex pentagons that tile fall into 15 known families. Karl Reinhardt found five in 1918; Richard Kershner three more in 1968; Richard James one in 1975; Marjorie Rice, an amateur mathematician working at her kitchen table, four more in 1976-77; Rolf Stein one in 1985; and Casey Mann, Jennifer McLoud-Mann and David Von Derau a fifteenth in 2015, by computer search. In 2017 Michaël Rao announced a computer-assisted proof that the list is complete.
Hexagons: convex hexagons that tile fall into exactly three families (Reinhardt, 1918).
Convex polygons with seven or more sides never tile the plane by themselves.

And a shape can tile in only strange ways. In the 1970s Roger Penrose found pairs of shapes that tile the plane but never periodically (no tiling by them repeats by sliding). Whether a single shape could do that, the "einstein problem" (German ein Stein, one stone), stayed open until 2023, when David Smith, Joseph Myers, Craig Kaplan and Chaim Goodman-Strauss found the "hat", a polygon made of eight kites that tiles the plane and only aperiodically. Chapter 10 returns to tilings through their symmetries: every periodic tiling has one of exactly 17 wallpaper groups.

3.5 Playfair's form of the fifth postulate and its equivalents¶

Level: intermediate to advanced.

Euclid's version and Playfair's are the same¶

Euclid's fifth postulate (Chapter 1) says: if a transversal makes co-interior angles adding to less than \(180°\) on one side, the two lines meet on that side. Playfair's says: through a point not on a line there is at most one parallel. They look different. In neutral geometry (all the axioms except the parallel one) they are equivalent: each implies the other.

We need one more neutral fact.

Lemma 3.15 (Euclid I.17). Any two angles of a triangle add up to less than \(180°\).

Proof. The exterior angle at \(C\) is \(180° - \angle C\), and by Theorem 2.2 it is larger than \(\angle B\). So \(\angle B < 180° - \angle C\), that is, \(\angle B + \angle C < 180°\). \(\blacksquare\)

Theorem 3.16. In neutral geometry, Euclid's fifth postulate and Playfair's axiom imply each other.

Proof that Euclid implies Playfair. Let \(P\) be a point not on \(\ell\). Drop the perpendicular \(PQ\) to \(\ell\), and let \(m\) be the line through \(P\) perpendicular to \(PQ\); it is parallel to \(\ell\) (Corollary 3.2). Let \(m'\) be any other line through \(P\). Then \(m'\) makes an angle less than \(90°\) with ray \(PQ\) on one side. On that side the co-interior angles at \(P\) and \(Q\) add to less than \(90° + 90° = 180°\), so by Euclid's postulate \(m'\) meets \(\ell\). So \(m\) is the only parallel. \(\blacksquare\)

Proof that Playfair implies Euclid. Let a transversal cross \(\ell\) at \(Q\) and \(m\) at \(P\), with co-interior angles \(\alpha\) (at \(P\)) and \(\beta\) (at \(Q\)) on one side, \(\alpha + \beta < 180°\). Through \(P\) draw the line \(m^*\) making co-interior angle \(180° - \beta\) with the transversal on that side; then the alternate angles are equal, so \(m^* \parallel \ell\) (Theorem 3.1). Since \(\alpha \ne 180° - \beta\), \(m \ne m^*\), so by Playfair \(m\) is not parallel to \(\ell\): they meet, at some point \(R\). Which side? If \(R\) were on the other side, triangle \(PQR\) would have angles \(180° - \alpha\) at \(P\) and \(180° - \beta\) at \(Q\), adding to \(360° - (\alpha + \beta) > 180°\), which Lemma 3.15 forbids. So \(R\) is on the side of the small angles, as Euclid says. \(\blacksquare\)

A long list of disguises¶

Over two thousand years, mathematicians trying to prove the fifth postulate kept producing statements that, on inspection, were the postulate in disguise: each is equivalent to it given the other axioms (for a few, including Archimedes' axiom that no length is infinitely small compared with another). Here are some, with the names usually attached to them.

Through a point not on a line there is exactly one parallel (Proclus in the 5th century; Playfair, 1795).
The angles of every triangle add to \(180°\).
There is at least one triangle whose angles add to \(180°\) (Saccheri and Legendre; this one needs Archimedes' axiom).
There exist two triangles that are similar but not congruent (John Wallis, 1663): the same shape in two sizes.
Every triangle has a circumscribed circle, a circle through its three vertices (Farkas Bolyai).
A rectangle exists (Clairaut, among others).
Parallel lines are everywhere the same distance apart; equivalently, the points at a fixed distance from a line, on one side, form a line (Proclus reports an attempt of this kind by Posidonius).
The Pythagorean theorem holds.
Through any point inside an angle there is a line meeting both arms (Legendre).

This chapter proved \(1 \Rightarrow 2\) (Theorem 3.8), \(1 \Rightarrow 7\) (Theorem 3.7), and with the remark on rectangles, \(1 \Rightarrow 6\). Chapter 4 proves \(1 \Rightarrow 8\), and Chapter 5 proves \(1 \Rightarrow 4\). The reverse directions are harder; Chapter 15 proves several of them.

Saccheri and the three hypotheses¶

The most determined attempt was by the Jesuit priest Girolamo Saccheri, in Euclides ab omni naevo vindicatus (Euclid freed of every flaw, 1733). Take a segment \(AB\), raise equal perpendiculars \(AD\) and \(BC\) on the same side, and join \(DC\). (Omar Khayyam had studied the same figure in the 11th century.) Using only neutral geometry, the two summit angles at \(C\) and \(D\) are equal (Exercise 10). There are three possibilities: they are right, obtuse or acute. Euclid's geometry is the hypothesis of the right angle. Saccheri showed, in neutral geometry, that whichever hypothesis holds for one such quadrilateral holds for all, and that it decides the angle sum of every triangle: equal to, more than, or less than \(180°\). He then refuted the obtuse hypothesis (it contradicts the infinite length of lines). For the acute hypothesis he derived theorem after strange theorem, found no contradiction, and finally declared it "repugnant to the nature of the straight line", which was not a proof. (In the figure the summit is drawn curved, to suggest the acute case; in that geometry the straight segment \(DC\) really does make acute angles with the sides.)

Johann Heinrich Lambert, in a work written in 1766 and published after his death, went further down the same road and noticed that under the acute hypothesis the angle deficit of a triangle is proportional to its area, just as the angle excess is on a sphere, as if the plane were a sphere of imaginary radius. He was very close. The final step, taken in the 1820s and 1830s by Gauss, János Bolyai and Nikolai Lobachevsky, was to stop looking for a contradiction and accept that there is none: the acute hypothesis gives a perfectly consistent geometry, the hyperbolic geometry. The parallel postulate cannot be proved because it is not true in every geometry that satisfies the other axioms. That story, and the proof that the new geometry is consistent, are the subject of Chapters 15 and 17.

Check yourself¶

1. Two parallel lines are cut by a transversal and one of the eight angles is 72°. What are the other seven?

Vertical angles and corresponding angles give three more of \(72°\), and their linear-pair partners are \(180° - 72° = 108°\). So four angles are \(72°\) and four are \(108°\).

2. Lines ℓ and m are cut by a transversal; a pair of corresponding angles are (4x + 9)° and (6x − 21)°. For which x are ℓ and m parallel, and what are the angles then?

By Theorems 3.1 and 3.3, the lines are parallel exactly when corresponding angles are equal: \(4x + 9 = 6x - 21\), \(x = 15\), and the angles are \(69°\) (and \(111°\)).

3. The angles of a triangle are x, 2x and 3x. Find them. What kind of triangle is it?

\(6x = 180°\), so \(x = 30°\), and the angles are \(30°, 60°, 90°\): a right triangle (the "30-60-90 triangle", half of an equilateral triangle).

4. In the bent-path figure, α = 35° and β = 48°. What is the angle at the bend? If instead the bend is a right angle and α is three times β, find α and β.

The bend is \(\alpha + \beta = 35° + 48° = 83°\). In the second case \(\alpha + \beta = 90°\) and \(\alpha = 3\beta\), so \(4\beta = 90°\), \(\beta = 22.5°\) and \(\alpha = 67.5°\).

5. What is the sum of the interior angles of a 12-gon? Each interior angle of a regular 12-gon?

\((12 - 2) \cdot 180° = 1800°\). Each angle of the regular one is \(1800/12 = 150°\) (exterior angle \(30°\)).

6. Each interior angle of a regular polygon is 162°. How many sides does it have, and how many diagonals?

Exterior angle \(180 - 162 = 18°\), so \(n = 360/18 = 20\). Diagonals: \(20 \cdot 17 / 2 = 170\).

7. Prove that the angles of any quadrilateral, convex or concave, add up to 360°.

By Lemma 3.11 a quadrilateral has a diagonal inside it. It cuts the quadrilateral into two triangles whose angles together make up the four angles of the quadrilateral. Each triangle contributes \(180°\) (Theorem 3.8), so the total is \(360°\). (For a concave quadrilateral, only the diagonal from the reflex vertex lies inside; the other one would not work.) \(\blacksquare\)

8. Prove that a parallelogram with one right angle is a rectangle, and that a parallelogram with equal diagonals is a rectangle.

Consecutive angles of a parallelogram add to \(180°\), so the neighbours of a right angle are right angles, and the opposite angle equals it: four right angles. For the second part: if \(ABCD\) is a parallelogram with \(AC = BD\), then triangles \(ABC\) and \(BAD\) have \(AB\) shared, \(BC = AD\) (opposite sides) and \(AC = BD\), so they are congruent by SSS and \(\angle ABC = \angle BAD\). These are consecutive angles, adding to \(180°\), so each is \(90°\). \(\blacksquare\)

9. Squares and regular octagons meet at a vertex as 4.8.8 (two of the three polygons are the same). Find every way to surround a vertex with three regular polygons, at least two of them alike.

Two \(n\)-gons and one \(m\)-gon fill the vertex when \(2\left(180 - \frac{360}{n}\right) + \left(180 - \frac{360}{m}\right) = 360\). Dividing by \(180\) and rearranging gives \(\frac{4}{n} + \frac{2}{m} = 1\). Try each \(n\): \(n = 3\) is impossible (\(4/3 > 1\)); \(n = 4\) needs \(2/m = 0\), impossible; \(n = 5\) gives \(m = 10\); \(n = 6\) gives \(m = 6\); \(n = 7\) gives \(m = 14/3\), no; \(n = 8\) gives \(m = 4\); \(n = 9\) gives \(m = 18/5\), no; \(n = 10\) gives \(m = 10/3\), no; \(n = 11\) gives \(m = 22/7\), no; \(n = 12\) gives \(m = 3\); and \(n \ge 13\) needs \(2/m > 9/13\), so \(m < 3\), impossible. The types are \(5.5.10\), \(6.6.6\), \(4.8.8\) and \(3.12.12\). Check \(3.12.12\): \(60 + 150 + 150 = 360\). Of these, \(6.6.6\), \(4.8.8\) and \(3.12.12\) extend to tilings; \(5.5.10\) does not (Exercise 11).

10. In a Saccheri quadrilateral ABCD (AD = BC, both perpendicular to AB), prove without the parallel postulate that the summit angles at C and D are equal.

Triangles \(DAB\) and \(CBA\) have \(DA = CB\), right angles at \(A\) and \(B\), and \(AB\) shared, so they are congruent by SAS, and \(DB = CA\). Then triangles \(ADC\) and \(BCD\) have \(AD = BC\), \(DC\) shared and \(AC = BD\), so they are congruent by SSS, and \(\angle ADC = \angle BCD\). \(\blacksquare\) (All the tools used, SAS and SSS, are neutral.)

11. Show that the vertex type 5.5.10 (two regular pentagons and a regular decagon) cannot be continued to a tiling of the plane.

Suppose every vertex is of type 5.5.10, and look at one pentagon \(T\). At each of its corners there is exactly one other pentagon and one decagon. The two edges of \(T\) at that corner are shared with these two neighbours, one each, so going around \(T\) the neighbours across its edges alternate: pentagon, decagon, pentagon, decagon, ... But \(T\) has five edges, an odd number, so the alternation cannot close up. Contradiction. \(\blacksquare\)

12. On Mars (radius 3389.5 km), a triangle of geodesic paths encloses 1000 km². By how much does its angle sum exceed 180°? How big would the triangle have to be for the excess to be 1°?

By Girard's theorem (Chapter 16) the excess is \(A / R^2 = 1000 / 3389.5^2 \approx 8.70 \times 10^{-5}\) radians, which is about \(0.0050°\), or \(18''\). For an excess of \(1° = \pi/180\) radians we need \(A = R^2 \pi / 180 \approx 3389.5^2 \times 0.017453 \approx 200\,500\ \text{km}^2\), about the area of a triangle with sides near 680 km.

What you learned¶

A transversal makes corresponding, alternate and co-interior angles; equal alternate angles force parallel lines in any geometry, but the converse needs the parallel postulate.
Playfair's axiom (at most one parallel through a point) is the form of the fifth postulate we use; it gives the properties of parallelograms and rectangles and the constant distance between parallel lines.
The angles of a triangle add to \(180°\); an exterior angle equals the sum of the two remote interior angles. Both fail on a sphere, where a triangle can have three right angles.
An \(n\)-gon's angles add to \((n - 2) \cdot 180°\), for concave polygons too, by triangulating; exterior angles of a convex polygon add to \(360°\); the tips of any five-pointed star add to \(180°\).
The equation \((k - 2)(n - 2) = 4\) shows that only triangles, squares and hexagons give regular tilings; there are eight semi-regular tilings; every triangle and every quadrilateral tiles, and in 2023 a single aperiodic tile was found.
Euclid's fifth postulate and Playfair's axiom are equivalent; many other statements (the angle sum, similar triangles, rectangles, Pythagoras) are disguised forms of it, and the failure to prove it led to hyperbolic geometry.

Next: Chapter 4, area: what it means, the formulas for the basic shapes proved by cutting and rearranging, and the theorem of Pythagoras, proved several ways.