Geometry

Dividing a segment into equal parts

Halving a segment takes two pairs of arcs. How do you cut a segment into three, five or seven exactly equal pieces when nothing can be measured?

You need

Triangle proportionality theoremGeometry Perpendicular and parallel lines through a pointGeometry

Where you are 8 in Compass and straightedge constructions

Geometry11

Compass and straightedge constructions8

▴ 3 above Perpendicular and parallel lines through a point Why a construction works Constructions with circles Equilateral triangle, square and hexagon in a circle

Dividing a segment into equal parts

When you reach for it

You have a segment \overline{AB} and you want the points that cut it into n equal pieces: three, five, seven, any whole number of them. Or you want the one point that splits it in a given ratio, such as 2:3. A ruler cannot do this exactly. A segment 10 cm long has its thirds at 3.333\ldots cm, which no scale shows. And bisecting, however often it is repeated, only reaches halves, quarters, eighths and so on, never a third.

The way round it is that equal pieces are easy to make somewhere else. A compass held at one width lays off equal steps along any line, as in copying a segment. The trouble is that those steps do not fit \overline{AB}: their total is whatever it happens to be. Parallel lines fix that. They carry the equal steps from a slanted ray across onto \overline{AB}, shrinking or stretching every one of them by the same factor, which is exactly what the triangle proportionality theorem says parallel lines do.

The tools are the usual two: a straightedge with no marks, and a compass.

The procedure

Dividing into n equal parts

The figure uses n = 5; nothing in the reasoning depends on it.

From A, draw a ray that leaves \overline{AB} at an angle. Any angle will do, as long as the ray does not lie along the line AB. Its job is to hold the equal steps, and once the last step is joined back to B it must make a triangle with \overline{AB}.

Open the compass to any width w and lay off n steps along the ray, end to end. With the point on A, mark A_1; with the point on A_1, mark A_2 beyond it; and so on up to A_n. Every mark is made with the same width, so

AA_1 = A_1A_2 = \cdots = A_{n-1}A_n = w

and so AA_k = kw for each k. The width has nothing to do with AB. It may be shorter or longer than a fifth of AB; it only has to be the same every time.

Draw \overline{A_nB}. This joins the last mark to the far end of the segment and closes the triangle AA_nB.

Through each of A_1, \ldots, A_{n-1}, draw the line parallel to \overline{A_nB}, and call the point where it meets \overline{AB} P_1, \ldots, P_{n-1}. To draw the parallel through A_k, copy \angle AA_nB so that its vertex is at A_k, one side runs back along the ray toward A, and the other opens on the same side as B. The new line and \overline{A_nB} then make equal corresponding angles with the ray, and equal corresponding angles force the lines to be parallel (proving lines parallel). This is the construction of a parallel through a point.

The points P_1, \ldots, P_{n-1} cut \overline{AB} into n equal parts.

Figure 1Five equal steps on the ray, the last one joined to B, and parallels through the others. The parallels carry the equal steps down onto \overline{AB}.

Why it works. Look at triangle AA_nB. The line through A_k and P_k crosses the sides \overline{AA_n} and \overline{AB} and is parallel to the third side, \overline{A_nB}. By the triangle proportionality theorem it cuts those two sides in the same ratio:

\frac{AP_k}{P_kB} = \frac{AA_k}{A_kA_n} = \frac{kw}{(n-k)w} = \frac{k}{n-k}

The width w cancels, and only the count of steps is left. So write AP_k = kt and P_kB = (n-k)t for some length t. Then AB = kt + (n-k)t = nt, so \displaystyle t = \frac{1}{n}AB and

AP_k = \frac{k}{n}\,AB

Writing P_0 = A and P_n = B, each piece is the difference of two of these:

P_{k-1}P_k = \frac{k}{n}\,AB - \frac{k-1}{n}\,AB = \frac{1}{n}\,AB

So all n pieces are equal, and they come in order along \overline{AB} because AP_k grows with k.

The same fact can be read through similar triangles. Triangles AA_kP_k and AA_nB share the angle at A, and their angles at A_k and A_n are corresponding angles on parallel lines, so the triangles are similar by AA. The scale factor is \displaystyle \frac{AA_k}{AA_n} = \frac{kw}{nw} = \frac{k}{n}, which gives \displaystyle AP_k = \frac{k}{n}AB at once, and also the length of each parallel cut, \displaystyle A_kP_k = \frac{k}{n}A_nB (Example 2).

Euclid uses this construction in Book VI, Proposition 9 of the Elements, to cut off a third of a line: three equal steps along a second line from the same end, the last joined to the far end, and a parallel through the first step.

Dividing in a given ratio

For a point P with AP:PB = m:n, where m and n are positive whole numbers, lay off m + n equal steps, join A_{m+n} to B, and draw just one parallel, through A_m. The same theorem gives

\frac{AP}{PB} = \frac{AA_m}{A_mA_{m+n}} = \frac{mw}{nw} = \frac{m}{n}

The steps are counted from A, so the first number of the ratio is the piece at A (Example 3).

One parallel instead of many

Drawing n - 1 parallels means copying an angle n - 1 times. There is a way to copy an angle only once. Draw the ray from A as before, and through B draw a ray parallel to it pointing the other way, on the other side of \overline{AB}: copy the angle between the first ray and \overline{AB} at B, so the two make equal alternate angles. With one compass width, mark A_1, \ldots, A_{n-1} along the ray from A and B_1, \ldots, B_{n-1} along the ray from B. Then join A_k to B_{n-k}, for each k, like the rungs of a ladder.

Let the rung \overline{A_kB_{n-k}} cross \overline{AB} at P. In triangles AA_kP and BB_{n-k}P the angles at A and B are equal alternate angles, because the rays are parallel and \overline{AB} crosses both (parallel lines and transversals), and the angles at P are vertical angles. So the triangles are similar by AA, and

\frac{AP}{PB} = \frac{AA_k}{BB_{n-k}} = \frac{kw}{(n-k)w} = \frac{k}{n-k}

That is the same ratio as before, so the rung crosses at P_k. Example 5 works it with numbers.

Worked examples

Example 1

Thirds of a 10 cm segment

\overline{AB} is 10 cm long. It is divided into three equal parts with the compass set to 4 cm for the steps along the ray. Find AA_3, the ratio in which each parallel cuts \overline{AB}, and how far each division point is from A.

Solution

Three steps of 4 cm give AA_3 = 3 \cdot 4 = 12 cm, which is longer than \overline{AB}. That does not matter: only the count of steps reaches the answer.

In triangle AA_3B the line \overleftrightarrow{A_1P_1} is parallel to \overline{A_3B}, so by the triangle proportionality theorem

\frac{AP_1}{P_1B} = \frac{AA_1}{A_1A_3} = \frac{4}{8} = \frac{1}{2}

So \overline{AB} is one part to two parts, three parts of 10 cm in all, and \displaystyle AP_1 = \frac{10}{3} \approx 3.33 cm. In the same way the parallel through A_2 gives

\frac{AP_2}{P_2B} = \frac{AA_2}{A_2A_3} = \frac{8}{4} = 2

two parts to one, so \displaystyle AP_2 = \frac{20}{3} \approx 6.67 cm.

Check. The three pieces are \displaystyle AP_1 = \frac{10}{3}, then \displaystyle P_1P_2 = \frac{20}{3} - \frac{10}{3} = \frac{10}{3}, and \displaystyle P_2B = 10 - \frac{20}{3} = \frac{10}{3}, all equal. No ruler shows \displaystyle \frac{10}{3} cm exactly; the construction lands there anyway.

Example 2

How long are the parallel cuts?

\overline{AB} is 12 cm long and is divided into four equal parts with steps of 2 cm, and the segment \overline{A_4B} turns out to be 9 cm. Find AP_k and the length of each parallel cut A_kP_k, for k = 1, 2, 3.

Solution

The ray holds AA_4 = 4 \cdot 2 = 8 cm. For each k, triangles AA_kP_k and AA_4B share the angle at A, and their angles at A_k and A_4 are corresponding angles on the parallels \overleftrightarrow{A_kP_k} and \overline{A_4B}. So they are similar by AA, with scale factor

\frac{AA_k}{AA_4} = \frac{2k}{8} = \frac{k}{4}

Every side of the small triangle is \displaystyle \frac{k}{4} of the matching side of the big one. On \overline{AB} that gives \displaystyle AP_k = \frac{k}{4}(12) = 3k, so AP_1 = 3, AP_2 = 6 and AP_3 = 9 cm. On the parallel side:

\begin{aligned} A_1P_1 &= \tfrac{1}{4}(9) = 2.25 \text{ cm} \\ A_2P_2 &= \tfrac{2}{4}(9) = 4.5 \text{ cm} \\ A_3P_3 &= \tfrac{3}{4}(9) = 6.75 \text{ cm} \end{aligned}

Check. The ratio \displaystyle \frac{AP_k}{AA_k} = \frac{3k}{2k} = \frac{3}{2} is the same for every k, and it equals \displaystyle \frac{AB}{AA_4} = \frac{12}{8}. The parallels stretch every step on the ray by the same factor \displaystyle \frac{3}{2}, which is why equal steps arrive as equal pieces.

Example 3

A point that splits a segment 2:3

\overline{AB} is 15 cm long. Construct the point P on it with AP:PB = 2:3, and find AP and PB.

Solution

The ratio has 2 + 3 = 5 parts, so lay off five equal steps along a ray from A, join A_5 to B, and draw one parallel to \overline{A_5B}, through A_2. It meets \overline{AB} at P.

Figure 2Example 3. Two steps to three, so AP:PB = 2:3, which on 15 cm is 6 cm and 9 cm.

In triangle AA_5B the line \overleftrightarrow{A_2P} is parallel to \overline{A_5B}, so

\frac{AP}{PB} = \frac{AA_2}{A_2A_5} = \frac{2w}{3w} = \frac{2}{3}

The five parts of 15 cm are 3 cm each, so AP = 2 \cdot 3 = 6 cm and PB = 3 \cdot 3 = 9 cm.

Check. 6 + 9 = 15 and \displaystyle \frac{6}{9} = \frac{2}{3}.

The order matters. The parallel through A_3 instead would give AP:PB = 3:2, a point 9 cm from A. The first number of the ratio is the count of steps from A.

Example 4

Checking the construction on a coordinate grid

A = (0, 0) and B = (12, 3). A ray from A through (1, 2) carries the marks A_k = (k, 2k). Show the steps are equal, then divide \overline{AB} into three equal parts: find the slope of \overline{A_3B}, the parallels through A_1 and A_2, and where they meet \overline{AB}.

Solution

From A to A_1 = (1, 2) is \displaystyle \sqrt{1^2 + 2^2} = \sqrt{5}, and each further mark adds (1, 2) again, so every step is \displaystyle \sqrt{5} long: one compass width. Three steps reach A_3 = (3, 6).

The slope of \overline{A_3B} is \displaystyle \frac{3 - 6}{12 - 3} = \frac{-3}{9} = -\frac{1}{3}, and the line AB is \displaystyle y = \frac{1}{4}x. Parallel lines have equal slopes, so the parallel through A_1 = (1, 2) is \displaystyle y = 2 - \frac{1}{3}(x - 1). It meets \overline{AB} where

\frac{x}{4} = 2 - \frac{x - 1}{3}

Multiplying by 12 gives 3x = 24 - 4x + 4, so 7x = 28 and x = 4. The point is P_1 = (4, 1).

The parallel through A_2 = (2, 4) is \displaystyle y = 4 - \frac{1}{3}(x - 2), and

\frac{x}{4} = 4 - \frac{x - 2}{3}

gives 3x = 48 - 4x + 8, so 7x = 56, x = 8 and P_2 = (8, 2).

Check. (4, 1) is \displaystyle \frac{1}{3} of the way from (0, 0) to (12, 3) and (8, 2) is \displaystyle \frac{2}{3} of the way. Each piece has length \displaystyle \sqrt{4^2 + 1^2} = \sqrt{17}, and three of them make \displaystyle AB = \sqrt{153} = 3\sqrt{17}. This is the answer partitioning a segment computes from coordinates; the construction reaches it with no coordinates at all.

Example 5

The ladder: one copied angle

\overline{AB} is 10 cm long and is to be cut into four equal parts by the ladder: a ray from A, a parallel ray from B pointing the other way, and steps of 2 cm on each, at A_1, A_2, A_3 and B_1, B_2, B_3. Show that the rung \overline{A_1B_3} crosses \overline{AB} 2.5 cm from A, and find where \overline{A_2B_2} and \overline{A_3B_1} cross it.

Solution

Let \overline{A_1B_3} cross \overline{AB} at P_1. In triangles AA_1P_1 and BB_3P_1, the angles at A and B are alternate angles made by \overline{AB} with the two parallel rays, so they are equal, and the angles at P_1 are vertical angles. So the triangles are similar by AA, and

\frac{AP_1}{P_1B} = \frac{AA_1}{BB_3} = \frac{2}{6} = \frac{1}{3}

Four parts of 10 cm: \displaystyle AP_1 = \frac{1}{4}(10) = 2.5 cm, and P_1B = 7.5 cm.

Figure 3Example 5. Two parallel rays pointing opposite ways, equal steps on each, and rungs joining A_k to B_{4-k}. Each rung crosses \overline{AB} at a division point.

The rung \overline{A_2B_2} has AA_2 = 4 and BB_2 = 4, a ratio of 1:1, so it crosses at the midpoint, 5 cm from A. The rung \overline{A_3B_1} has AA_3 = 6 and BB_1 = 2, a ratio of 3:1, so it crosses \displaystyle \frac{3}{4}(10) = 7.5 cm from A.

The three crossings, at 2.5, 5 and 7.5 cm, are the quarter points. Only one angle was copied, the one that made the second ray parallel to the first; the rungs come out parallel to each other on their own (Exercise 22).

Example 6

Locating \displaystyle \frac{5}{7} on a number line

A number line has 0 and 1 marked 10 cm apart. Construct the point \displaystyle \frac{5}{7} and find its distance from 0. Then construct \displaystyle \frac{12}{7}.

Solution

Treat the segment from 0 to 1 as \overline{AB}. Lay off seven equal steps along a ray from 0, join A_7 to the point 1, and draw the parallel through A_5. It meets the number line \displaystyle \frac{5}{7} of the way from 0 to 1. That point is the number \displaystyle \frac{5}{7}, since the segment from 0 to 1 is the unit. On this drawing it sits

\frac{5}{7}(10) = \frac{50}{7} \approx 7.14 \text{ cm}

from 0. Repeated halving could never reach it, because halving only makes denominators that are powers of 2, and 7 is not one.

For \displaystyle \frac{12}{7} = 1 + \frac{5}{7}, set the compass to the distance from 0 to \displaystyle \frac{5}{7} and copy it along the line starting at 1. The new point is \displaystyle 10 + \frac{50}{7} = \frac{120}{7} \approx 17.14 cm from 0. This is how a fraction is located on the number line with the unit fraction \displaystyle \frac{1}{7} made exact rather than estimated.

Boundaries

The ray must leave the line AB. If the steps are laid along the line AB itself, A_n is on that line, there is no triangle AA_nB, and the parallels to \overline{A_nB} through the marks are that same line, so they cross \overline{AB} nowhere in particular. Any other direction works.

Equal steps, end to end, starting at A. One compass width for every step, each starting where the last ended, all on the one ray and the first starting at A. The ratio \displaystyle \frac{k}{n-k} came from the steps being equal; unequal steps divide \overline{AB} in their own unequal ratio (Exercise 21).

Truly parallel. Each parallel is constructed by copying an angle. A line slid across by eye is a measurement in disguise, and the reasoning only holds for exact parallels.

Whole numbers of steps, or lengths in hand. The construction divides in any ratio m:n of positive whole numbers. For a ratio a:b of two given lengths, lay off a segment of length a and then one of length b along the ray in place of the equal steps; the same theorem gives AP:PB = a:b. Both parts of the ratio have to be on the paper, as a count of steps or as drawn lengths: a ratio given only as a number with no length to match it is out of reach until that length is constructed.

The ladder needs parallel rays pointing opposite ways, and one width on both. If the two rays are not parallel, the alternate angles are not equal and the triangles are not similar.

Exact in principle. The reasoning is exact; a drawing is as exact as the pencil. The lengths in the examples say where the construction lands. A straightedge has no marks, and nothing is measured.

Where this goes next

Partitioning a segment in a ratio does on coordinates what this construction does on paper: the point \displaystyle \frac{k}{n} of the way from A to B, found by taking \displaystyle \frac{k}{n} of each coordinate change, as in Example 4. The case n = 2 is the midsegment fact turned round: the line through the midpoint of one side of a triangle, parallel to a second side, meets the third side at its midpoint.

The two similar triangles AA_kP_k and AA_nB are related by a dilation with center A and scale factor \displaystyle \frac{k}{n}, which carries A_n to A_k and B to P_k. So this construction draws a dilation with compass and straightedge alone. The same picture, with the steps replaced by two given lengths, builds a length that completes a proportion: lay off a and then b on one ray, c on another, join and draw a parallel, and the new piece is \displaystyle \frac{bc}{a}. And the argument used here, naming the equal steps the compass made and the parallel the copied angle made, then reading off the ratio, is the pattern for justifying any construction.

Exercises

Equal parts

\overline{AB} is 14 cm long and is divided into five equal parts. How far from A is each of the four division points?
\overline{AB} is 9 cm long and is divided into four equal parts with steps of 3 cm along the ray. Find AA_4 and AP_3.
\overline{AB} is divided into six equal parts, and \overline{A_6B} is 15 cm long. Find the parallel cuts A_2P_2 and A_5P_5.
\overline{AB} is divided into three equal parts, and AP_1 = 4.2 cm. Find AB and AP_2.
\overline{AB} is 21 cm long and is divided into seven equal parts with steps of 2.5 cm. Find AA_7, the factor \displaystyle \frac{AB}{AA_7} by which the parallels stretch each step, and AP_4.

In a given ratio

\overline{AB} is 20 cm long. To construct P with AP:PB = 3:2, how many steps are laid off, which mark gets the parallel, and what are AP and PB?
\overline{AB} is 18 cm long. Do the same for AP:PB = 4:5.
\overline{AB} is 28 cm long. Do the same for AP:PB = 2:5.
\overline{AB} is 9 cm long. Do the same for a point Q with AQ:QB = 5:1.

On a coordinate grid

A = (0, 0) and B = (10, 0), and the ray carries the marks A_k = (k, k) for k = 1 to 5. Find the equation of the line A_5B, the parallel through A_2, and the point where it meets \overline{AB}.
A = (0, 0) and B = (6, 6), and the ray along the x-axis carries the marks A_k = (k, 0) for k = 1, 2, 3. Find the two points that divide \overline{AB} into three equal parts, using the parallels to \overline{A_3B}.
A = (1, 2) and B = (13, 8). The construction divides \overline{AB} into three equal parts. Where do the two division points land?

Fractions and the ladder

On a number line, 0 and 1 are 14 cm apart. To locate \displaystyle \frac{3}{7}, how many steps are laid off, which mark gets the parallel, and how far from 0 does the point land?
On a number line, 0 and 1 are 6 cm apart. How many steps and which mark locate \displaystyle \frac{4}{9}, and how far from 0 does it land?
On a number line, 0 and 1 are 6 cm apart. Describe how to construct \displaystyle \frac{7}{3} and say how far from 0 it lands.
\overline{AB} is 12 cm long and is cut into three equal parts by the ladder, with steps of 2 cm. How far from A does the rung \overline{A_1B_2} cross \overline{AB}?
\overline{AB} is 15 cm long and is cut into five equal parts by the ladder. The rung \overline{A_2B_3} crosses \overline{AB} at P. Find AP:PB and AP.

Reasoning

Explain why the compass width used for the steps along the ray can be anything at all.
Using similar triangles rather than the triangle proportionality theorem, prove that \displaystyle AP_k = \frac{k}{n}AB, and deduce that every piece is \displaystyle \frac{1}{n}AB.
Explain what goes wrong if the ray is drawn along the line AB.
\overline{AB} is 14 cm long. Someone lays off steps of 2 cm, 2 cm and 3 cm along the ray, joins the last mark to B, and draws parallels through the other two. Explain why the three pieces of \overline{AB} are not equal, and find them.
In the ladder, prove that the rungs \overline{A_kB_{n-k}} are all parallel to each other.

Answers

2.8 cm, 5.6 cm, 8.4 cm and 11.2 cm. Each part is \displaystyle \frac{14}{5} = 2.8 cm, and \displaystyle AP_k = \frac{k}{5}(14).
AA_4 = 12 cm and AP_3 = 6.75 cm. Four steps of 3 cm make 12 cm, and \displaystyle AP_3 = \frac{3}{4}(9). The steps are longer than the parts, which changes nothing.
A_2P_2 = 5 cm and A_5P_5 = 12.5 cm. Triangle AA_kP_k is similar to triangle AA_6B with scale factor \displaystyle \frac{k}{6}, so \displaystyle A_kP_k = \frac{k}{6}(15).
AB = 12.6 cm and AP_2 = 8.4 cm. AP_1 is one third of AB, so AB = 3(4.2), and AP_2 is two thirds, 2(4.2).
AA_7 = 17.5 cm, the factor is \displaystyle \frac{21}{17.5} = \frac{6}{5}, and AP_4 = 12 cm. AA_4 = 4(2.5) = 10 and \displaystyle \frac{6}{5}(10) = 12, the same as \displaystyle \frac{4}{7}(21).
Five steps, the parallel through A_3; AP = 12 cm and PB = 8 cm. Each of the five parts is \displaystyle \frac{20}{5} = 4 cm.
Nine steps, the parallel through A_4; AP = 8 cm and PB = 10 cm. Each part is \displaystyle \frac{18}{9} = 2 cm.
Seven steps, the parallel through A_2; AP = 8 cm and PB = 20 cm. Each part is \displaystyle \frac{28}{7} = 4 cm.
Six steps, the parallel through A_5; AQ = 7.5 cm and QB = 1.5 cm. Each part is \displaystyle \frac{9}{6} = 1.5 cm.
y = -x + 10, then y = -x + 4, meeting \overline{AB} at (4, 0). The slope of \overline{A_5B} is \displaystyle \frac{0 - 5}{10 - 5} = -1 and it passes through (10, 0). The parallel through A_2 = (2, 2) has slope -1 too, and it meets the x-axis at x = 4, which is \displaystyle \frac{2}{5} of the way from A.
(2, 2) and (4, 4). A_3 = (3, 0), and the slope of \overline{A_3B} is \displaystyle \frac{6 - 0}{6 - 3} = 2. The parallel through (1, 0) is y = 2x - 2, which meets y = x at x = 2; the parallel through (2, 0) is y = 2x - 4, which meets it at x = 4.
(5, 4) and (9, 6). The change from A to B is (12, 6); a third of it is (4, 2), so the points are (1 + 4, 2 + 2) and (1 + 8, 2 + 4).
Seven steps, the parallel through A_3, landing 6 cm from 0, since \displaystyle \frac{3}{7}(14) = 6.
Nine steps, the parallel through A_4, landing \displaystyle \frac{8}{3} \approx 2.67 cm from 0, since \displaystyle \frac{4}{9}(6) = \frac{24}{9} = \frac{8}{3}.
14 cm from 0. Divide the segment from 0 to 1 into three equal parts, each 2 cm, then copy that third with the compass seven times along the line from 0: 7(2) = 14, which is \displaystyle \frac{7}{3}(6).
4 cm from A. Triangles AA_1P and BB_2P are similar with AA_1 = 2 and BB_2 = 4, so AP:PB = 1:2 and \displaystyle AP = \frac{1}{3}(12).
AP:PB = 2:3 and AP = 6 cm. Triangles AA_2P and BB_3P are similar with AA_2:BB_3 = 2:3, and \displaystyle \frac{2}{5}(15) = 6.
Every ratio in the argument is a ratio of steps, \displaystyle \frac{AA_k}{A_kA_n} = \frac{kw}{(n-k)w}, and the width w cancels to leave \displaystyle \frac{k}{n-k}. A different width gives a different triangle AA_nB and different parallels, but the same points P_k on \overline{AB}.
Triangles AA_kP_k and AA_nB share \angle A, and \angle AA_kP_k = \angle AA_nB as corresponding angles, since \overleftrightarrow{A_kP_k} is parallel to \overline{A_nB}. So they are similar by AA with scale factor \displaystyle \frac{AA_k}{AA_n} = \frac{kw}{nw} = \frac{k}{n}, and the matching sides give \displaystyle AP_k = \frac{k}{n}AB. Then each piece is \displaystyle AP_k - AP_{k-1} = \frac{k}{n}AB - \frac{k-1}{n}AB = \frac{1}{n}AB, with P_0 = A and P_n = B.
The mark A_n lands on the line AB, so A, A_n and B make no triangle. The line through A_n and B is the line AB itself, and a parallel to it through a mark on it is that same line again, so it never crosses \overline{AB} at a single point.
4 cm, 4 cm and 6 cm. The last mark A_3 is 2 + 2 + 3 = 7 cm from A. The parallel through A_1 cuts AP_1:P_1B = 2:5, so \displaystyle AP_1 = \frac{2}{7}(14) = 4; the one through A_2 cuts 4:3, so \displaystyle AP_2 = \frac{4}{7}(14) = 8. The parallels copy the pattern of the steps, 2:2:3, onto \overline{AB}, and unequal steps give unequal pieces.
Steps \overline{A_kA_{k+1}} and \overline{B_{n-k-1}B_{n-k}} are both one compass width long and lie on parallel rays, so in the quadrilateral A_kA_{k+1}B_{n-k-1}B_{n-k} one pair of opposite sides is equal and parallel. That makes it a parallelogram (proving a quadrilateral is a parallelogram), so its other two sides, the rungs \overline{A_kB_{n-k}} and \overline{A_{k+1}B_{n-k-1}}, are parallel. Going along, every rung is parallel to the next, so all of them are parallel.