---
title: "Thales’ Theorem"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 68
exercises: 9
source: https://one-course.com/books/math/1/en/chapter/68-thales-theorem
---

# Chapter 68 — Thales’ Theorem

How do you measure the height of a [pyramid](https://one-course.com/books/math/1/en/chapter/62-pyramids-and-cones#def-g8-solids-def) without climbing it? Thales of Miletus compared its shadow with the shadow of a stick. His theorem — [parallel lines](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) cut [segments](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) in [proportional](https://one-course.com/books/math/1/en/chapter/67-linear-and-affine-functions#def-g9-linfunc-linear) pieces — is the mathematical heart of [scale](https://one-course.com/books/math/1/en/chapter/49-proportionality#def-g7-prop-scale) models, maps, and enlargements, and one of the oldest theorems with a name.

## 68.1 The theorem

**Theorem 68.1 (Thales).**

Let two lines meet at a point $A$. Take $M$ on the first line and $N$ on the second, with $B$ further along the first and $C$ further along the second. If the lines $(MN)$ and $(BC)$ are *[parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)*, then the triangles $AMN$ and $ABC$ have [proportional](https://one-course.com/books/math/1/en/chapter/67-linear-and-affine-functions#def-g9-linfunc-linear) sides:

$$
\frac{AM}{AB} = \frac{AN}{AC} = \frac{MN}{BC}.
$$

**Proof.** *Admitted at this level.* ∎

![The two Thales configurations: nested triangles (left) and the “butterfly”, where M and N sit on the other side of A (right). In both, (MN) (BC) and the three ratios are equal.](https://one-course.com/images/onecourse/chapters/math-1/g9-thales/fig-45201f71a40b.svg)

![The two Thales configurations: nested triangles (left) and the “butterfly”, where M and N sit on the other side of A (right). In both, (MN) (BC) and the three ratios are equal.](https://one-course.com/images/onecourse/chapters/math-1/g9-thales/fig-9ab852c1ceba.svg)

*The two Thales configurations: nested triangles (left) and the “butterfly”, where $M$ and $N$ sit on the other side of $A$ (right). In both, $(MN) \parallel (BC)$ and the three ratios are equal.*

**Method 68.2 (Computing a length with Thales).**

1. Identify the point $A$ where the two lines cross and the two [parallel lines](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) ; name the two triangles;
2. write the three equal ratios, always “small triangle over large triangle”;
3. keep the equality of the two ratios involving the three known lengths and the unknown one;
4. solve the resulting proportion by cross-multiplication.

**Example 68.3.**

In the left configuration above, suppose $AM = 3$, $AB = 5$, $AN = 2.4$ and $MN = 3.3$, and let us compute $AC$ and $BC$. Thales gives

$$
\frac{AM}{AB} = \frac{AN}{AC} = \frac{MN}{BC},
\qquad\text{i.e.}\qquad
\frac35 = \frac{2.4}{AC} = \frac{3.3}{BC}.
$$

From $\frac35 = \frac{2.4}{AC}$, cross-multiplying: $3 \times AC = 5 \times 2.4 = 12$, so $AC = 4$. From $\frac35 = \frac{3.3}{BC}$: $3 \times BC = 16.5$, so $BC = 5.5$.

**Example 68.4 (Thales in real life).**

A vertical stick of height $1$ m casts a shadow of $1.5$ m, while a tree casts a shadow of $12$ m at the same moment. Sun rays are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp), so the stick-and-shadow and tree-and-shadow triangles are in Thales configuration:

$$
\frac{\text{height of tree}}{1} = \frac{12}{1.5},
\qquad\text{so the tree is } 8 \text{ m tall.}
$$

## 68.2 The converse

**Theorem 68.5 (Converse of Thales).**

With the points placed as in [Theorem 68.1](#thm-g9-thales-direct) ($M$ on $(AB)$, $N$ on $(AC)$, in the same order on both lines): if

$$
\frac{AM}{AB} = \frac{AN}{AC},
$$

then the lines $(MN)$ and $(BC)$ are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp).

**Proof.** *Admitted at this level.* ∎

**Method 68.6 (Proving or disproving parallelism).**

1. Compute separately the two ratios $\dfrac{AM}{AB}$ and $\dfrac{AN}{AC}$ (as [fractions](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-fraction) , not [roundings](https://one-course.com/books/math/1/en/chapter/38-decimal-numbers#def-g6-decimals-rounding) );
2. if they are equal *and* the points are in the same order on the two lines, the lines are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) (converse of Thales);
3. if they differ, the lines are *not* [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) — because if they were, Thales’ theorem would force the ratios to be equal.

**Example 68.7.**

On line 1, $AM = 4$ and $AB = 10$; on line 2, $AN = 6$ and $AC = 15$. Then

$$
\frac{AM}{AB} = \frac{4}{10} = \frac25,
\qquad
\frac{AN}{AC} = \frac{6}{15} = \frac25 :
$$

equal ratios, same order: $(MN) \parallel (BC)$.

With $AC = 14$ instead: $\frac{6}{14} = \frac37 \neq \frac25$, so the lines are not [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp).

## 68.3 Enlargement and reduction

**Definition 68.8 (Scaling a figure).**

Enlarging or reducing a figure by the *scale factor* $k > 0$ means multiplying *all* its lengths by $k$: an *enlargement* when $k > 1$, a *reduction* when $k < 1$. In Thales’ configuration, the triangle $AMN$ is the reduction of $ABC$ by the factor $k = \frac{AM}{AB}$.

**Proposition 68.9 (Effect on angles and areas).**

Scaling by the factor $k$ preserves angles and the shape of figures, multiplies every length by $k$, and multiplies every *[area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area)* by $k^2$.

**Proof.** *Admitted at this level.* ∎

![Doubling the lengths (k = 2) multiplies the area by k2 = 4: four copies of the small rectangle tile the large one.](https://one-course.com/images/onecourse/chapters/math-1/g9-thales/fig-13be5c7be8eb.svg)

*Doubling the lengths ($k = 2$) multiplies the [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) by $k^2 = 4$: four copies of the small rectangle tile the large one.*

**Example 68.10.**

A photo of $10 \times 15$ cm is enlarged with [scale factor](#def-g9-thales-scaling) $k = 3$: the print measures $30 \times 45$ cm. Its [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) goes from $150$ cm$^2$ to $150 \times 9 = 1350$ cm$^2$ — $9$ times larger, not $3$.

## 68.4 Exercises

**Exercise 68.1 ★.**

The lines $(MN)$ and $(BC)$ are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp), with $M$ on $[AB]$ and $N$ on $[AC]$; $AM = 2$, $AB = 6$, $AN = 3$, $BC = 9$. Compute $AC$ and $MN$.

**Solution of Exercise 68.1.**

Thales: $\dfrac{AM}{AB} = \dfrac{AN}{AC} = \dfrac{MN}{BC}$, i.e. $\dfrac26 = \dfrac13$. So $\dfrac{3}{AC} = \dfrac13$ gives $AC = 9$, and $\dfrac{MN}{9} = \dfrac13$ gives $MN = 3$.

**Exercise 68.2 ★.**

Same configuration: $AM = 5$, $MB = 3$ (careful: $MB$, not $AB$!), $AN = 4$. Compute $AC$.

**Solution of Exercise 68.2.**

First find $AB = AM + MB = 5 + 3 = 8$. Then $\dfrac{AM}{AB} = \dfrac{AN}{AC}$ gives $\dfrac58 = \dfrac{4}{AC}$, so $5 \times AC = 32$ and $AC = 6.4$.

**Exercise 68.3 ★.**

In a butterfly configuration, $A$ is between $M$ and $B$ and between $N$ and $C$, with $(MN) \parallel (BC)$; $AM = 3$, $AB = 7.5$, $AN = 2$, $MN = 2.6$. Compute $AC$ and $BC$.

**Solution of Exercise 68.3.**

The butterfly works exactly like the nested configuration: $\dfrac{AM}{AB} = \dfrac{AN}{AC} = \dfrac{MN}{BC}$, i.e. $\dfrac{3}{7.5} = \dfrac25$. So $\dfrac{2}{AC} = \dfrac25$ gives $AC = 5$, and $\dfrac{2.6}{BC} = \dfrac25$ gives $BC = \dfrac{2.6 \times 5}{2} = 6.5$.

**Exercise 68.4 ★.**

$M$ is on $[AB]$ with $AM = 6$, $AB = 8$; $N$ is on $[AC]$ with $AN = 9$, $AC = 12$. Are the lines $(MN)$ and $(BC)$ [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)?

**Solution of Exercise 68.4.**

$\dfrac{AM}{AB} = \dfrac68 = \dfrac34$ and $\dfrac{AN}{AC} = \dfrac{9}{12} = \dfrac34$: equal ratios, points in the same order, so $(MN) \parallel (BC)$ by the converse of Thales.

**Exercise 68.5 ★★.**

Same as above with $AM = 4$, $AB = 6$, $AN = 5$, $AC = 8$. Are $(MN)$ and $(BC)$ [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)? Justify carefully.

**Solution of Exercise 68.5.**

$\dfrac{AM}{AB} = \dfrac46 = \dfrac23$ and $\dfrac{AN}{AC} = \dfrac58$. Cross-check: $\dfrac23 = \dfrac{16}{24}$ and $\dfrac58 = \dfrac{15}{24}$: the ratios differ. If the lines were [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp), Thales’ theorem would force them to be equal — so $(MN)$ and $(BC)$ are *not* [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp).

**Exercise 68.6 ★★.**

A $1.8$ m tall person stands $2$ m away from a street lamp’s base; their shadow measures $3$ m. Draw the Thales configuration formed by the lamp, the person, and the tip of the shadow, and compute the height of the lamp.

**Solution of Exercise 68.6.**

Let $S$ be the tip of the shadow, $P$ the top of the person’s head, $L$ the top of the lamp. The person ($1.8$ m at distance $3$ m from $S$) and the lamp (height $h$ at distance $3 + 2 = 5$ m from $S$) are two [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) vertical [segments](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) cut by the light ray $(SL)$: Thales from the point $S$ gives

$$
\frac{1.8}{h} = \frac{3}{5},
\qquad\text{so}\quad
h = \frac{1.8 \times 5}{3} = 3 \text{ m}.
$$

**Exercise 68.7 ★★.**

A map has [scale](https://one-course.com/books/math/1/en/chapter/49-proportionality#def-g7-prop-scale) $1 : 25\,000$ (lengths on the map are the real lengths multiplied by $k = \frac{1}{25000}$).

1. Two villages are $6.8$ cm apart on the map. What is the real distance, in km?
2. A forest has an [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) of $8$ cm $^2$ on the map. What is its real [area](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) , in km $^2$ ?

**Solution of Exercise 68.7.**

*1.* Real distance: $6.8 \times 25\,000 = 170\,000$ cm $=
1.7$ km.

*2.* [Areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) scale by $k^2$: real area $= 8 \times 25\,000^2$ cm$^2$ $= 8 \times 6.25 \times 10^8$ cm$^2$ $= 5 \times 10^9$ cm$^2$. Since $1$ km$^2 = 10^{10}$ cm$^2$, that is $0.5$ km$^2$.

**Exercise 68.8 ★★.**

Triangle $ABC$ has $AB = 12$, $AC = 15$, $BC = 18$. The point $M$ on $[AB]$ satisfies $AM = 8$, and the line through $M$ [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(BC)$ cuts $[AC]$ at $N$.

1. Compute $AN$ and $MN$ .
2. What is the [scale factor](#def-g9-thales-scaling) from $ABC$ to $AMN$ ? Compare the [perimeters](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) of the two triangles.

**Solution of Exercise 68.8.**

*1.* The ratio is $\dfrac{AM}{AB} = \dfrac{8}{12} = \dfrac23$. Thales: $AN = \frac23 \times 15 = 10$ and $MN = \frac23 \times 18 = 12$.

*2.* The scale factor is $k = \frac23$. Perimeters: $ABC$ has $12 + 15 + 18 = 45$, and $AMN$ has $8 + 10 + 12 = 30 = \frac23 \times
45$: the perimeter scales by $k$ too, like every length.

**Exercise 68.9 ★★★.**

Let $ABCD$ be a trapezoid with $(AB) \parallel (CD)$, $AB = 4$ and $CD = 6$, whose diagonals $[AC]$ and $[BD]$ meet at $O$. Using Thales in the butterfly configuration around $O$, compute the ratio $\dfrac{OA}{OC}$, and show that $O$ cuts both diagonals in the same ratio.

**Solution of Exercise 68.9.**

Around $O$, the lines $(AC)$ and $(BD)$ cross, and $(AB) \parallel (CD)$: butterfly configuration with the triangles $OAB$ and $OCD$. Thales gives

$$
\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD} = \frac46 = \frac23 .
$$

So $\dfrac{OA}{OC} = \dfrac23$, and the equality $\dfrac{OA}{OC} = \dfrac{OB}{OD}$ says precisely that $O$ cuts the two diagonals in the same ratio.

## 68.5 Problem: The unmarked ruler, the pinhole camera, and the three means of a trapezoid

**Problem 68.1.**

Weekend problem — Thales at work: dividing any segment into equal parts, measuring the Sun with a shoebox, and a trapezoid where all three famous means meet

Thales’ theorem is the mathematics of *rays*: sun rays, light rays through a pinhole, pencil rays from a [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def). This problem uses it three ways — as a construction tool (dividing a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) of *any* length, [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) $\sqrt2$, into perfectly equal parts), as a measuring instrument (the [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) of the Sun, with a shoebox and a coin), and as a magnifying glass on the trapezoid of [Exercise 68.9](#exo-g9-thales-9), where the arithmetic, geometric and harmonic means of this series all turn up in one figure.

**Part I — Dividing with an unmarked ruler.**

1. Draw a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[AB]$ of $7$ cm. To cut it into three equal parts with compass and unmarked ruler: draw any ray from $A$ (not through $B$ ); step off three equal compass lengths on it, giving points $P_1$ , $P_2$ , $P_3$ ; join $P_3$ to $B$ ; draw the [parallels](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $(P_3 B)$ through $P_1$ and $P_2$ . Perform the construction and mark where the [parallels](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) cut $[AB]$ .
2. Justify with [Theorem 68.1](#thm-g9-thales-direct) that the two marked points cut $[AB]$ exactly at its third points.
3. Adapt the method to divide a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) into $5$ equal parts, then to construct $\frac35$ of a given [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) .
4. Construct the point $M$ of $[AB]$ with $\frac{AM}{MB} = \frac23$ (that is, $M$ at $\frac25$ of the way from $A$ ). How many equal steps on the ray, and which point joins $B$ ?
5. Why is this construction better than measuring with a graduated ruler? Consider a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) of length $\sqrt2$ (a unit square’s diagonal, irrational by [Problem 65.1](https://one-course.com/books/math/1/en/chapter/65-square-roots#pb-g9-sqrt-1) ): what would measuring give, and what does Thales give?

**Part II — The pinhole camera.** Poke a pin through one [face](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) of a closed [box](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def): on the opposite [face](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def), an upside-down image of the world appears. A point of an object, the pinhole, and the image point are aligned — light travels straight — so the object (height $H$, at distance $D$ in front of the hole) and its image (height $h$, on the back wall at depth $d$ behind the hole) sit in the butterfly configuration of Thales.

6. Derive the pinhole formula $h = H \times \frac dD$ from Thales in the butterfly around the hole.
7. A $12$ m tree stands $30$ m from a shoebox of depth $20$ cm. How tall is its image, and which way up?
8. Point the [box](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) at the Sun: at $d = 1$ m behind the pinhole, the Sun’s image is a disk of [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) about $9$ mm. Given the Sun’s distance $D = 1.5 \times 10^{11}$ m, compute its [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) in [scientific notation](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#def-g9-fractions-scientific) ( [Problem 63.1](https://one-course.com/books/math/1/en/chapter/63-fractions-and-powers#pb-g9-fractions-1) ’s notation at work). Compare with the true value, $1.39 \times 10^9$ m.
9. The Moon: [diameter](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-circle) $3.5 \times 10^6$ m, distance $3.8 \times 10^8$ m. Compute the ratio $\frac{\text{diameter}}{\text{distance}}$ for the Moon and for the Sun. What lucky coincidence do the two numbers reveal — and what spectacular event does it make possible?
10. In one or two sentences: why is the pinhole image upside down, and why does enlarging the pinhole make the image brighter but blurrier?

**Part III — The trapezoid of the three means.** $ABCD$ is the trapezoid of [Exercise 68.9](#exo-g9-thales-9): $(AB) \parallel (CD)$, $AB = 4$, $CD = 6$, diagonals crossing at $O$ with $\frac{OA}{OC} = \frac{OB}{OD} = \frac{4}{6} =
\frac23$.

11. Butterfly practice: two lines cross at $O$ with $(MN) \parallel (PQ)$ in butterfly position; $OM = 3$ , $OP = 7.5$ , $MN = 4$ and $OQ = 6$ . Compute $PQ$ and $ON$ .
12. Explain how Thales, applied with ratio $\frac12$ , contains the midpoint theorem of [Theorem 59.1](https://one-course.com/books/math/1/en/chapter/59-midpoints-and-parallels#thm-g8-midpoints-direct) as its special case.
13. Draw the [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to the bases through $O$ ; it meets $[AD]$ at $E$ . Working in the triangle $ACD$ (note $\frac{AO}{AC} = \frac25$ ), compute $EO$ .
14. By the mirror computation in the triangle $BDC$, the piece $OF$ (with $F$ on $[BC]$) has the same length. Conclude that $$EF = 2 \times \frac{4 \times 6}{4 + 6} = 4.8 :$$ the [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) through the diagonal crossing measures the *[harmonic mean](https://one-course.com/books/math/1/en/chapter/61-proportionality-speed-and-averages#pb-g8-speed-1)* of the bases — the round-trip mean of [Problem 61.1](https://one-course.com/books/math/1/en/chapter/61-proportionality-speed-and-averages#pb-g8-speed-1), reappearing in pure geometry.
15. The grand finale: compute the three classical means of the bases $4$ and $6$ — arithmetic $\frac{4+6}{2}$ , geometric $\sqrt{4 \times 6}$ , harmonic $\frac{2 \times 4 \times 6}{4 + 6}$ — and order them. Each lives in the trapezoid: the arithmetic mean is the midline joining the legs’ midpoints ( [Problem 53.1](https://one-course.com/books/math/1/en/chapter/53-areas-and-volumes#pb-g7-areas-1) ), the [harmonic mean](https://one-course.com/books/math/1/en/chapter/61-proportionality-speed-and-averages#pb-g8-speed-1) is your [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $EF$ , and the [geometric mean](https://one-course.com/books/math/1/en/chapter/60-the-right-triangle-and-the-cosine#pb-g8-cosine-1) is the [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) that cuts the trapezoid into two *similar* trapezoids (check its value with a calculator; the proof is a lovely extra challenge). Where has the inequality between the three been proved in this book ( [Problem 60.1](https://one-course.com/books/math/1/en/chapter/60-the-right-triangle-and-the-cosine#pb-g8-cosine-1) , [Problem 61.1](https://one-course.com/books/math/1/en/chapter/61-proportionality-speed-and-averages#pb-g8-speed-1) )?

**Solution of Problem 68.1.**

**1.** The construction produces two points on $[AB]$; call them $M_1$ (from $P_1$) and $M_2$ (from $P_2$).

**2.** In the triangle $A P_3 B$, the parallels to $(P_3 B)$ through $P_1$ and $P_2$ cut the sides proportionally ([Theorem 68.1](#thm-g9-thales-direct)):

$$
\frac{A M_1}{AB} = \frac{A P_1}{A P_3} = \frac13,
\qquad
\frac{A M_2}{AB} = \frac{A P_2}{A P_3} = \frac23 .
$$

The compass made $A P_1$, $P_1 P_2$, $P_2 P_3$ equal, so the ratios are exactly thirds — whatever the angle of the ray and the compass opening chosen.

**3.** For fifths: step off five equal lengths, join the fifth point to $B$, and draw four parallels. For $\frac35$ of a segment: same figure, and take the point cut by the parallel through $P_3$: it sits at $\frac{3}{5}$ of $[AB]$ from $A$.

**4.** $\frac{AM}{MB} = \frac23$ means $AM = \frac25 AB$: five equal steps on the ray, join $P_5$ to $B$, and the parallel through $P_2$ marks $M$.

**5.** Measuring $\sqrt2 = 1.41421\ldots$ cm only ever uses finitely many decimals: any measured “third” is an approximation. The Thales construction never reads a number: it delivers the exact point at $\frac{\sqrt2}{3}$ — equal parts of a segment no ruler can even express ([Problem 65.1](https://one-course.com/books/math/1/en/chapter/65-square-roots#pb-g9-sqrt-1)).

**6.** The object’s top, the hole and the image’s bottom are aligned, and likewise for the object’s bottom and the image’s top: two lines crossing at the hole, with the object and the image wall parallel. Thales in the butterfly:

$$
\frac{h}{H} = \frac{d}{D},
\qquad\text{so}\qquad
h = H \times \frac dD .
$$

**7.** $h = 12 \times \frac{0.20}{30} = 0.08$ m $= 8$ cm — upside down (the rays cross at the hole).

**8.** $H = h \times \frac Dd = 9 \times 10^{-3} \times
\frac{1.5 \times 10^{11}}{1} = 1.35 \times 10^{9}$ m. True value $1.39 \times 10^9$ m: a shoebox measures the Sun to within a few percent.

**9.** Moon: $\frac{3.5 \times 10^6}{3.8 \times 10^8}
\approx 0.0092$. Sun: $\frac{1.39 \times 10^9}{1.5 \times
10^{11}} \approx 0.0093$. The two ratios — the apparent sizes in the sky — are almost identical: the Moon can cover the Sun *exactly*, rim to rim. That cosmic fluke is the total solar eclipse.

**10.** Every ray must pass through the one hole, so rays from the top of the object continue *down* and rays from the bottom continue *up*: the image is inverted. Enlarging the hole lets through a whole bundle of slightly shifted copies of the image, which overlap: brighter, but smeared.

**11.** $\frac{PQ}{MN} = \frac{OP}{OM} = \frac{7.5}{3}
= 2.5$, so $PQ = 10$; and $\frac{ON}{OQ} = \frac{OM}{OP} =
\frac{1}{2.5}$, so $ON = \frac{6}{2.5} = 2.4$.

**12.** If the parallel passes through the midpoint of one side, the Thales ratio is $\frac12$, so it cuts the second side at *its* midpoint, and the parallel segment measures half the base: precisely [Theorem 59.1](https://one-course.com/books/math/1/en/chapter/59-midpoints-and-parallels#thm-g8-midpoints-direct). Thales is the midpoint theorem freed from the ratio $\frac12$.

**13.** In the triangle $ACD$, the line $(EO)$ is parallel to the base $(CD)$, with $\frac{AO}{AC} = \frac{OA}{OA + OC} =
\frac{4}{4 + 6} = \frac25$. Thales: $\frac{EO}{DC} = \frac{AO}{AC} = \frac25$, so $EO = \frac25 \times 6 = 2.4$.

**14.** In the triangle $BDC$, likewise $\frac{BO}{BD} = \frac25$ and $OF = \frac25 \times 6 = 2.4$. Hence

$$
EF = EO + OF = 4.8
= \frac{2 \times 4 \times 6}{4 + 6} :
$$

the harmonic mean of the bases — the mean of round trips ([Problem 61.1](https://one-course.com/books/math/1/en/chapter/61-proportionality-speed-and-averages#pb-g8-speed-1)), drawn in a trapezoid.

**15.** Arithmetic: $5$; geometric: $\sqrt{24} \approx 4.90$; harmonic: $4.8$. Order: $4.8 < 4.90\ldots < 5$, i.e. harmonic $<$ geometric $<$ arithmetic — with equality only for equal bases. In the figure: the midline measures $5$, the parallel through the diagonal crossing $4.8$, and the similarity cut $\sqrt{24}$, all stacked between the bases in that order. The inequalities were proved in [Problem 60.1](https://one-course.com/books/math/1/en/chapter/60-the-right-triangle-and-the-cosine#pb-g8-cosine-1) (geometric $\leq$ arithmetic, by circle and by identity) and [Problem 61.1](https://one-course.com/books/math/1/en/chapter/61-proportionality-speed-and-averages#pb-g8-speed-1) (harmonic $\leq$ arithmetic); harmonic $\leq$ geometric follows by combining them ($H \times A = G^2$, proved there too).