A bundle of parallel lines cuts two transversals into proportional parts. It is with this idea that a board is divided into equal parts without doing any math.

Who discovered

The result is called Tales of Miletus (6th century BC), associated by tradition with the measurement of the height of the Great Pyramid by shadows (see the curiosity about it at the opening of the chapter). The demonstration we know is from Euclid, in proposition VI.2 of The Elements (c. 300 BC), and uses areas of triangles. The name “Tales’ theorem” for this result is a tradition in countries such as Brazil, Portugal, France, Spain and Italy; in German and English, the same name refers to the angle inscribed in the semicircle.

Thales of Miletus, later artistic representation.
Thales of Miletus, later artistic representation.Luca Giordano · Wikimedia Commons · Public domain
Euclid, Renaissance representation traditionally associated with Euclid or Archimedes.
Euclid, Renaissance representation traditionally associated with Euclid or Archimedes.Rafael Sanzio · Wikimedia Commons · Public domain

The story

There is no record that Thales demonstrated the theorem. What the sources say is that he used proportionality to measure heights and distances. Euclid’s proof also needed to work for incommensurable segments, and that is why it is based on the theory of proportions of Eudoxus, from book V.

The name seems to have stuck in French textbooks at the end of the 19th century and from there it passed into Brazilian teaching. The attribution of the angle in the semicircle to Thales is older and comes from Diógenes Laércio.

Diógenes Laércio, later artistic representation.
Diógenes Laércio, later artistic representation.Unidentified writer · Wikimedia Commons · Public domain

In everyday life

  • Divide a board into equal parts. The carpenter tilts the tape measure over the board until a “round” measurement (e.g. 30 cm to 5 parts) fits from edge to edge, and marks from 6 to 6 cm: the marks divide the width into equal parts.
  • Scales. Maps, models and plans maintain the proportions of the original.
  • Staves and grids. Equally spaced parallel lines cut any inclined line into equal parts.
  • Perspective. Designers use proportionality to space posts and windows that move apart in the drawing.

Sources: Tales (MacTutor) · Elements, proposition VI.2 · Elements, book V.

Theory and constructions

DefiniçãoParallel and transversal beam

  • Beam of parallel lines is a set of coplanar lines parallel to each other.
  • Transverse of the beam is a straight line in the plane that cuts all the straight lines of the beam.
  • Corresponding points of two transversals are on the same straight line of the beam; corresponding segments have endpoints at corresponding points.

PropriedadeCongruent parts

If two straight lines are transversals of a bundle of distinct parallels and a segment of one of them is divided into $p$ congruent parts, and straight lines of the bundle pass through the division points, then the corresponding segment of the other transversal is also divided into $p$ parts congruent to each other.

View the demo

1st part (number of parts). If the corresponding segment were divided into fewer (or more) parts, at least two straight lines of the beam would meet — absurd, as they are parallel.

2nd part (congruence). Through the division points of the second transversal we draw parallels to the first. Parallelograms and triangles are formed with a side congruent to $x$ and angles with respectively parallel sides; by ALA, the triangles are all congruent. Therefore, the parts of the second transversal are congruent.

TeoremaThales' theorem

If two lines are transversals of a bundle of parallel lines, then the ratio between any two segments of one of them is equal to the ratio between the respective corresponding segments of the other:

$$\frac{AB}{CD} = \frac{A^{\prime}B^{\prime}}{C^{\prime}D^{\prime}}, \qquad\text{ou ainda}\qquad \frac{AB}{A^{\prime}B^{\prime}} = \frac{CD}{C^{\prime}D^{\prime}}.$$

The ratio between corresponding segments is constant.

Drag the parallel ones (purple balls on the left) and the transversal ones (gold balls). The ratios between corresponding segments remain the same.
View the demo

1st case: $\overline{AB}$ and $\overline{CD}$ commensurable. There is a segment $x$ that fits $p$ times in $\overline{AB}$ and $q$ times in $\overline{CD}$: $\dfrac{AB}{CD} = \dfrac{p}{q}$. Drawing beam straight lines through the division points, using the previous property $A^{\prime}B^{\prime} = p\,x^{\prime}$ and $C^{\prime}D^{\prime} = q\,x^{\prime}$, then $\dfrac{A^{\prime}B^{\prime}}{C^{\prime}D^{\prime}} = \dfrac{p}{q}$.

2nd case: incommensurable. We take $y$ that fits $n$ times in $\overline{CD}$; then $m\,y < AB < (m+1)\,y$ to some $m$, that is, $\dfrac{m}{n} < \dfrac{AB}{CD} < \dfrac{m+1}{n}$. According to the previous property, the same goes for $\dfrac{A^{\prime}B^{\prime}}{C^{\prime}D^{\prime}}$. By decreasing $y$ (increasing $n$), the fractions $\dfrac{m}{n}$ and $\dfrac{m+1}{n}$ define a single real number — which is equal to both ratios.

ExemploDivide a segment into equal parts

To divide $\overline{AB}$ into $p$ congruent parts, we draw an auxiliary half-line through $A$, mark $p$ equal segments on it with the compass, connect the last mark to $B$ and draw parallel to this line through the other marks. By Thales’ theorem, they divide $\overline{AB}$ into $p$ congruent parts.

Construction with ruler and compass: change the number of parts and drag $B$ and the auxiliary ray.

PropriedadeParallel to one side of the triangle

A straight line parallel to one side of a triangle that cuts the other two sides divides them in the same ratio. Conversely, if $\dfrac{AD}{DB} = \dfrac{AE}{EC}$, with $D \in \overline{AB}$ and $E \in \overline{AC}$, then $\overline{DE} \parallel \overline{BC}$ (exercise 427).