Thales's theorem

[[File:Dante-thales-theorem.jpg|thumb|right|upright=1.2|

{{lang|la|Non si est dare primum motum esse}}

{{lang|it|o se del mezzo cerchio far si puote

triangol sì c'un recto nonauesse.}}

– Dante's Paradiso, Canto 13, lines 100–102

{{lang|la|Non si est dare primum motum esse,}}

Or if in semicircle can be made

Triangle so that it have no right angle.

– English translation by Longfellow

Babylonian mathematicians knew this for special cases before Greek mathematicians proved it.de Laet, Siegfried J. (1996). History of Humanity: Scientific and Cultural Development. UNESCO, Volume 3, p. 14. {{isbn|92-3-102812-X}}

Thales of Miletus (early 6th century BC) is traditionally credited with proving the theorem; however, even by the 5th century BC there was nothing extant of Thales' writing, and inventions and ideas were attributed to men of wisdom such as Thales and Pythagoras by later doxographers based on hearsay and speculation.{{cite journal |last=Dicks |first=D. R. |title=Thales |journal=The Classical Quarterly |year=1959 |volume=9 |number=2 |pages=294–309 |doi=10.1017/S0009838800041586 }}{{cite web |first=G. Donald |last=Allen |title=Thales of Miletus |url=http://www.math.tamu.edu/~dallen/masters/Greek/thales.pdf |date=2000 |access-date=2012-02-12}} Reference to Thales was made by Proclus (5th century AD), and by Diogenes Laërtius (3rd century AD) documenting Pamphila's (1st century AD) statement that Thales "was the first to inscribe in a circle a right-angle triangle".{{cite journal |last1=Patronis |first1=Tasos |last2=Patsopoulos |first2=Dimitris |date=January 2006 |title=The Theorem of Thales: A Study of the Naming of Theorems in School Geometry Textbooks |journal=The International Journal for the History of Mathematics Education |issn=1932-8826 |archive-url=https://web.archive.org/web/20131105213425/http://journals.tc-library.org/index.php/hist_math_ed/article/viewFile/189/184 |url=http://journals.tc-library.org/index.php/hist_math_ed/article/viewFile/189/184 |url-status=usurped |archive-date=2013-11-05 |pages=57–68}}

Thales was claimed to have traveled to Egypt and Babylonia, where he is supposed to have learned about geometry and astronomy and thence brought their knowledge to the Greeks, along the way inventing the concept of geometric proof and proving various geometric theorems. However, there is no direct evidence for any of these claims, and they were most likely invented speculative rationalizations. Modern scholars believe that Greek deductive geometry as found in Euclid's Elements was not developed until the 4th century BC, and any geometric knowledge Thales may have had would have been observational.{{r|dicks}}{{cite book |last=Sidoli |first=Nathan |year=2018 |chapter=Greek mathematics |editor1-last=Jones |editor1-first=A. |editor2-last=Taub |editor2-first=L. |title=The Cambridge History of Science: Vol. 1, Ancient Science |publisher=Cambridge University Press |pages=345–373 |chapter-url=http://individual.utoronto.ca/acephalous/Sidoli_2018c.pdf }}

The theorem appears in Book III of Euclid's Elements ({{c.|300 BC}}) as proposition 31: "In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; further the angle of the greater segment is greater than a right angle, and the angle of the less segment is less than a right angle."

Dante Alighieri's Paradiso (canto 13, lines 101–102) refers to Thales's theorem in the course of a speech.

The following facts are used: the sum of the angles in a triangle is equal to 180° and the base angles of an isosceles triangle are equal.

{{Gallery

|File:Animated illustration of thales theorem.gif|Provided {{mvar|{{overline|AC}}}} is a diameter, angle at {{mvar|B}} is constant right (90°).

|File:Thales' Theorem.svg|Figure for the proof.

}}

Since {{math|1={{overline|OA}} = {{overline|OB}} = {{overline|OC}}}}, {{math|△OBA}} and {{math|△OBC}} are isosceles triangles, and by the equality of the base angles of an isosceles triangle, {{math|1=∠ OBC = ∠ OCB}} and {{math|1=∠ OBA = ∠ OAB}}.

Let {{math|1=α = ∠ BAO}} and {{math|1=β = ∠ OBC}}. The three internal angles of the {{math|∆ABC}} triangle are {{mvar|α}}, {{math|(α + β)}}, and {{mvar|β}}. Since the sum of the angles of a triangle is equal to 180°, we have

:$\backslash begin\{align\}$

\alpha+ (\alpha + \beta) + \beta &= 180^\circ \\

2\alpha + 2\beta &= 180^\circ \\

2( \alpha + \beta ) &= 180^\circ \\

\therefore \alpha + \beta &= 90^\circ.

\end{align}

Q.E.D.

The theorem may also be proven using trigonometry: Let {{math|1=O = (0, 0)}}, {{math|1=A = (−1, 0)}}, and {{math|1=C = (1, 0)}}. Then {{mvar|B}} is a point on the unit circle {{math|(cos θ, sin θ)}}. We will show that {{math|△ABC}} forms a right angle by proving that {{mvar|{{overline|AB}}}} and {{mvar|{{overline|BC}}}} are perpendicular — that is, the product of their slopes is equal to −1. We calculate the slopes for {{mvar|{{overline|AB}}}} and {{mvar|{{overline|BC}}}}:

:$\backslash begin\{align\}$

m_{AB} &= \frac{y_B - y_A}{x_B - x_A} = \frac{\sin \theta}{\cos \theta + 1} \\[2pt]

m_{BC} &= \frac{y_C - y_B}{x_C - x_B} = \frac{-\sin \theta}{-\cos \theta + 1}

\end{align}

Then we show that their product equals −1:

: $\backslash begin\{align\}$

&m_{AB} \cdot m_{BC}\\[4pt]

= {} & \frac{\sin \theta}{\cos \theta + 1} \cdot \frac{-\sin \theta}{-\cos \theta + 1}\\[4pt]

= {} & \frac{-\sin ^2 \theta}{-\cos ^2 \theta +1}\\[4pt]

= {} & \frac{-\sin ^2 \theta}{\sin ^2 \theta}\\[4pt]

= {} & {-1}

\end{align}

Note the use of the Pythagorean trigonometric identity $\backslash sin^2\; \backslash theta\; +\; \backslash cos^2\; \backslash theta\; =\; 1.$

File:Thales_theorem_by_refelection1.svg

Let {{math|△ABC}} be a triangle in a circle where {{mvar|AB}} is a diameter in that circle. Then construct a new triangle {{math|△ABD}} by mirroring {{math|△ABC}} over the line {{mvar|AB}} and then mirroring it again over the line perpendicular to {{mvar|AB}} which goes through the center of the circle. Since lines {{mvar|AC}} and {{mvar|BD}} are parallel, likewise for {{mvar|AD}} and {{mvar|CB}}, the quadrilateral {{mvar|ACBD}} is a parallelogram. Since lines {{mvar|AB}} and {{mvar|CD}}, the diagonals of the parallelogram, are both diameters of the circle and therefore have equal length, the parallelogram must be a rectangle. All angles in a rectangle are right angles.

The theorem can be proved using vector algebra. Let's take the vectors $\backslash overrightarrow\{AB\}$ and $\backslash overrightarrow\{CB\}$. These vectors satisfy

:$\backslash overrightarrow\{AB\}\; =\; \backslash overrightarrow\{AO\}\; +\; \backslash overrightarrow\{OB\}\backslash qquad\; \backslash qquad\; \backslash overrightarrow\{CB\}\; =\; \backslash overrightarrow\{CO\}\; +\; \backslash overrightarrow\{OB\}$

and their dot product can be expanded as

:$\backslash overrightarrow\{AB\}\backslash cdot\backslash overrightarrow\{CB\}\; =\; \backslash left(\backslash overrightarrow\{AO\}\; +\; \backslash overrightarrow\{OB\}\backslash right)\backslash cdot\backslash left(\; \backslash overrightarrow\{CO\}\; +\; \backslash overrightarrow\{OB\}\backslash right)\; =\backslash overrightarrow\{AO\}\backslash cdot\backslash overrightarrow\{CO\}+(\backslash overrightarrow\{AO\}+\backslash overrightarrow\{CO\})\backslash cdot\; \backslash overrightarrow\{OB\}+\backslash overrightarrow\{OB\}\backslash cdot\backslash overrightarrow\{OB\}$

but

:$\backslash overrightarrow\{AO\}=-\backslash overrightarrow\{CO\}\backslash qquad\backslash qquad\; \backslash overrightarrow\{AO\}\backslash cdot\backslash overrightarrow\{CO\}=-R^2\backslash qquad\backslash qquad\; \backslash overrightarrow\{OB\}\backslash cdot\backslash overrightarrow\{OB\}\; =\; R^2$

and the dot product vanishes

:$\backslash overrightarrow\{AB\}\backslash cdot\backslash overrightarrow\{CB\}\; =\; -R^2+\backslash overrightarrow\{0\}\backslash cdot\; \backslash overrightarrow\{OB\}+R^2=0$

and then the vectors $\backslash overrightarrow\{AB\}$ and $\backslash overrightarrow\{CB\}$ are orthogonal and the angle ABC is a right angle.

For any triangle, and, in particular, any right triangle, there is exactly one circle containing all three vertices of the triangle. This circle is called the circumcircle of the triangle.

{{collapse top|title=Uniqueness proof (sketch)}}

The locus of points equidistant from two given points is a straight line that is called the perpendicular bisector of the line segment connecting the points. The perpendicular bisectors of any two sides of a triangle intersect in exactly one point. This point must be equidistant from the vertices of the triangle.

{{collapse bottom}}

One way of formulating Thales's theorem is: if the center of a triangle's circumcircle lies on the triangle then the triangle is right, and the center of its circumcircle lies on its hypotenuse.

The converse of Thales's theorem is then: the center of the circumcircle of a right triangle lies on its hypotenuse. (Equivalently, a right triangle's hypotenuse is a diameter of its circumcircle.)

Image:Thales' Theorem Converse.svg

This proof consists of 'completing' the right triangle to form a rectangle and noticing that the center of that rectangle is equidistant from the vertices and so is the center of the circumscribing circle of the original triangle, it utilizes two facts:

• adjacent angles in a parallelogram are supplementary (add to 180°) and,
• the diagonals of a rectangle are equal and cross each other in their median point.

Let there be a right angle {{math|∠ ABC}}, {{mvar|r}} a line parallel to {{mvar|{{overline|BC}}}} passing by {{mvar|A}}, and {{mvar|s}} a line parallel to {{mvar|{{overline|AB}}}} passing by {{mvar|C}}. Let {{mvar|D}} be the point of intersection of lines {{mvar|r}} and {{mvar|s}}. (It has not been proven that {{mvar|D}} lies on the circle.)

The quadrilateral {{mvar|ABCD}} forms a parallelogram by construction (as opposite sides are parallel). Since in a parallelogram adjacent angles are supplementary (add to 180°) and {{math|∠ ABC}} is a right angle (90°) then angles {{math|∠ BAD, ∠ BCD, ∠ ADC}} are also right (90°); consequently {{mvar|ABCD}} is a rectangle.

Let {{mvar|O}} be the point of intersection of the diagonals {{mvar|{{overline|AC}}}} and {{overline|BD}}. Then the point {{mvar|O}}, by the second fact above, is equidistant from {{mvar|A}}, {{mvar|B}}, and {{mvar|C}}. And so {{mvar|O}} is center of the circumscribing circle, and the hypotenuse of the triangle ({{mvar|{{overline|AC}}}}) is a diameter of the circle.

Given a right triangle {{mvar|ABC}} with hypotenuse {{mvar|AC}}, construct a circle {{math|Ω}} whose diameter is {{mvar|AC}}. Let {{mvar|O}} be the center of {{math|Ω}}. Let {{mvar|D}} be the intersection of {{math|Ω}} and the ray {{mvar|OB}}. By Thales's theorem, {{math|∠ ADC}} is right. But then {{mvar|D}} must equal {{mvar|B}}. (If {{mvar|D}} lies inside {{math|△ABC}}, {{math|∠ ADC}} would be obtuse, and if {{mvar|D}} lies outside {{math|△ABC}}, {{math|∠ ADC}} would be acute.)

This proof utilizes two facts:

• two lines form a right angle if and only if the dot product of their directional vectors is zero, and
• the square of the length of a vector is given by the dot product of the vector with itself.

Let there be a right angle {{math|∠ ABC}} and circle {{mvar|M}} with {{overline|AC|}} as a diameter.

Let M's center lie on the origin, for easier calculation.

Then we know

• {{math|1=A = −C}}, because the circle centered at the origin has {{mvar|{{overline|AC}}}} as diameter, and
• {{math|1=(A – B) · (B – C) = 0}}, because {{math|∠ ABC}} is a right angle.

It follows

:$\backslash begin\{align\}$

0 &= (A-B) \cdot (B-C) \\

&= (A-B) \cdot (B+A) \\

&= |A|^2 - |B|^2. \\[4pt]

\therefore \ |A| &= |B|.

\end{align}

This means that {{mvar|A}} and {{mvar|B}} are equidistant from the origin, i.e. from the center of {{mvar|M}}. Since {{mvar|A}} lies on {{mvar|M}}, so does {{mvar|B}}, and the circle {{mvar|M}} is therefore the triangle's circumcircle.

The above calculations in fact establish that both directions of Thales's theorem are valid in any inner product space.

As stated above, Thales's theorem is a special case of the inscribed angle theorem (the proof of which is quite similar to the first proof of Thales's theorem given above):

:Given three points {{mvar|A}}, {{mvar|B}} and {{mvar|C}} on a circle with center {{mvar|O}}, the angle {{math|∠ AOC}} is twice as large as the angle {{math|∠ ABC}}.

A related result to Thales's theorem is the following:

• If {{mvar|{{overline|AC}}}} is a diameter of a circle, then:

:*If {{mvar|B}} is inside the circle, then {{math|∠ ABC > 90°}}

:*If {{mvar|B}} is on the circle, then {{math|1=∠ ABC = 90°}}

:*If {{mvar|B}} is outside the circle, then {{math|∠ ABC < 90°}}.

Image:Thales' Theorem Tangents.svg

Thales's theorem can be used to construct the tangent to a given circle that passes through a given point. In the figure at right, given circle {{mvar|k}} with centre {{mvar|O}} and the point {{mvar|P}} outside {{mvar|k}}, bisect {{mvar|{{overline|OP}}}} at {{mvar|H}} and draw the circle of radius {{mvar|{{overline|OH}}}} with centre {{mvar|H}}. {{mvar|{{overline|OP}}}} is a diameter of this circle, so the triangles connecting OP to the points {{mvar|T}} and {{mvar|T′}} where the circles intersect are both right triangles.

File:Root_construction_geometric_mean5.svg $h=\backslash sqrt\{pq\}$ with $q=1$]]

Thales's theorem can also be used to find the centre of a circle using an object with a right angle, such as a set square or rectangular sheet of paper larger than the circle.[https://books.google.com/books?id=9jASBwAAQBAJ&pg=PA183 Resources for Teaching Mathematics: 14–16] Colin Foster The angle is placed anywhere on its circumference (figure 1). The intersections of the two sides with the circumference define a diameter (figure 2). Repeating this with a different set of intersections yields another diameter (figure 3). The centre is at the intersection of the diameters.

File:Thales_theorem_find_circle_centre.svg

• Synthetic geometry
• Inverse Pythagorean theorem

{{reflist}}

• {{cite book |last1=Agricola |first1=Ilka |author1-link=Ilka Agricola |last2=Friedrich |first2=Thomas |year=2008 |title=Elementary Geometry |publisher=AMS |isbn=978-0-8218-4347-5 |page=50}}
• {{cite book |last=Heath |first=T.L. |author-link=Thomas L. Heath |year=1921 |title=A History of Greek Mathematics: From Thales to Euclid |volume=I |publisher=Oxford |pages=131ff |url=https://archive.org/details/historyofgreekma01heat/page/130/}}

• {{MathWorld |title=Thales' Theorem |urlname=ThalesTheorem}}
• [http://www.cut-the-knot.org/pythagoras/Munching/inscribed.shtml Munching on Inscribed Angles]
• [http://www.mathopenref.com/thalestheorem.html Thales's theorem explained], with interactive animation
• [http://demonstrations.wolfram.com/ThalesTheorem/ Demos of Thales's theorem] by Michael Schreiber, The Wolfram Demonstrations Project.

{{Ancient Greek mathematics}}

Category:Euclidean plane geometry

Category:Greek mathematics

Category:Articles containing proofs

Category:Theorems about right triangles

Category:Theorems about triangles and circles

es:Teorema de Tales#Segundo teorema

he:משפט תאלס#המשפט השני