Two figures are similar when they have the same shape, even if they are different sizes. It’s the mathematics of photos, maps and models.

Who discovered

The idea is in the stories about Tales of Miletus, who would have measured the height of a pyramid by comparing its shadow with that of a stick. The complete theory is in book VI of The Elements by Euclid (c. 300 BC), which defines similar figures as those with equal angles and proportional sides. It depends on the theory of proportions by Eudoxus of Cnidus (4th century BC), from book V, which also allowed the treatment of ratios between incommensurable segments.

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

The shadow method became an instrument. From the Middle Ages onwards, astrolabes and quadrants were engraved with the shadow square, with two scales, the umbra recta and the umbra versa, to measure the heights of towers and the depths of wells without complicated calculations.

Old engraving: the height of a tower is measured with the square of the shadows (umbra recta and umbra versa), from two positions of the observer.
Old engraving: the height of a tower is measured with the square of the shadows (umbra recta and umbra versa), from two positions of the observer.Collection of the Euler’s Eyes blog.

Around 270 B.C., Aristarchus of Samos used similar triangles to estimate the relative distances from the Earth to the Moon and the Sun. The numbers were far from the real ones, but the method—comparing similar triangles in the sky—was correct.

Aristarchus of Samos, later artistic representation.
Aristarchus of Samos, later artistic representation.Claudio Camozzi · Wikimedia Commons · Public domain

In everyday life

  • Enlarge a photo. When “zooming” without distorting, all measurements are multiplied by the same factor.
  • Mockups and miniatures. A cart in scale 1 : 43 has each measurement 43 times smaller than that of the real car.
  • Height of a building by shadow. Compare the shadow of the building with that of a person of known height, at the same time.
  • Projector and camera. The image on the screen or sensor is similar to the object, with size proportional to the distance.

Sources: Tales (MacTutor) · Elements, book VI · Aristarchus (MacTutor).

Theory and constructions

DefiniçãoSimilar triangles

Two triangles are similar (symbol $\sim$) if, and only if, they have three orderly congruent angles and proportional homologous sides:

$$\triangle ABC \sim \triangle A^{\prime}B^{\prime}C^{\prime} \iff \begin{cases} \hat{A} \equiv \hat{A}^{\prime},\ \hat{B} \equiv \hat{B}^{\prime},\ \hat{C} \equiv \hat{C}^{\prime} \\ \dfrac{a}{a^{\prime}} = \dfrac{b}{b^{\prime}} = \dfrac{c}{c^{\prime}} = k \end{cases}$$

homologous sides (homo = same, logos = place) are one in each triangle and are opposite congruent angles. The number $k$ is the similarity ratio; if $k = 1$, the triangles are congruent.

Similarity is reflexive, symmetrical and transitive.

Change the ratio $k$ and rotate the copy: equal angles, proportional sides. Perimeters and heights are also in the ratio $k$; the areas, in the ratio $k^2$.

ExemploCalculating with the similarity ratio

If $\triangle ABC \sim \triangle A^{\prime}B^{\prime}C^{\prime}$, with $A^{\prime}B^{\prime} = 3$ cm, $A^{\prime}C^{\prime} = 7$ cm, $B^{\prime}C^{\prime} = 5$ cm and $AB = 6$ cm, the ratio is $k = \dfrac{6}{3} = 2$. So $BC = 2 \cdot 5 = 10$ cm and $AC = 2 \cdot 7 = 14$ cm.

TeoremaFundamental theorem

If a straight line is parallel to one of the sides of a triangle and intersects the other two at different points, then the triangle it determines is similar to the first:

$$\overline{DE} \parallel \overline{BC} \Rightarrow \triangle ADE \sim \triangle ABC.$$

Demo

  1. Angles: such as $\overline{DE} \parallel \overline{BC}$, $\hat{D} \equiv \hat{B}$ and $\hat{E} \equiv \hat{C}$ (corresponding), and $\hat{A}$ is common.
  2. Sides: by Thales’ theorem, $\dfrac{AD}{AB} = \dfrac{AE}{AC}$.
  3. By $E$ we draw $\overline{EF} \parallel \overline{AB}$, with $F$ in $\overline{BC}$. $BDEF$ is parallelogram, so $DE = BF$, and by Tales $\dfrac{AE}{AC} = \dfrac{BF}{BC} = \dfrac{DE}{BC}$.
  4. Therefore $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$, and the triangles are similar.
Drag $D$: parallel $\overline{DE}$ always cuts a triangle similar to the original.

ExemploApplying the fundamental theorem

In a triangle with $AB = 12$ cm, $AC = 13$ cm and $BC = 15$ cm, the parallel $\overline{DE}$ to $\overline{BC}$ is $DE = 5$ cm. Then $\dfrac{x}{12} = \dfrac{y}{13} = \dfrac{5}{15}$, that is, $AD = x = 4$ cm and $AE = y = \dfrac{13}{3}$ cm.