Thales's theorem

Thales’ theorem: if Template:Mvar is a diameter and Template:Mvar is a point on the diameter's circle, the angle Template:Math is a right angle.

In geometry, Thales's theorem states that if Template:Mvar, Template:Mvar, and Template:Mvar are distinct points on a circle where the line Template:Mvar is a diameter, the angle Template:Math is a right angle. Thales's theorem is a special case of the inscribed angle theorem and is mentioned and proved as part of the 31st proposition in the third book of Euclid's Elements.<ref>Template:Cite book Originally published by Cambridge University Press. 1st edition 1908, 2nd edition 1926. </ref> It is generally attributed to Thales of Miletus, but it is sometimes attributed to Pythagoras.

HistoryEdit

File:Dante-thales-theorem.jpg

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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.<ref>de Laet, Siegfried J. (1996). History of Humanity: Scientific and Cultural Development. UNESCO, Volume 3, p. 14. Template:Isbn</ref>

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.<ref name=dicks>Template:Cite journal</ref><ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> 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".<ref>Template:Cite journal</ref>

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.Template:R<ref>Template:Cite book</ref>

The theorem appears in Book III of Euclid's Elements (Template:C.) 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.

ProofEdit

First proofEdit

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. {{#invoke:Gallery|gallery}}

Since Template:Math, Template:Math and Template:Math are isosceles triangles, and by the equality of the base angles of an isosceles triangle, Template:Math and Template:Math.

Let Template:Math and Template:Math. The three internal angles of the Template:Math triangle are Template:Mvar, Template:Math, and Template:Mvar. Since the sum of the angles of a triangle is equal to 180°, we have

<math>\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}</math>

Q.E.D.

Second proofEdit

The theorem may also be proven using trigonometry: Let Template:Math, Template:Math, and Template:Math. Then Template:Mvar is a point on the unit circle Template:Math. We will show that Template:Math forms a right angle by proving that Template:Mvar and Template:Mvar are perpendicular — that is, the product of their slopes is equal to −1. We calculate the slopes for Template:Mvar and Template:Mvar:

<math>\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}</math>

Then we show that their product equals −1:

<math>\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}</math>

Note the use of the Pythagorean trigonometric identity <math>\sin^2 \theta + \cos^2 \theta = 1.</math>

Third proofEdit

File:Thales theorem by refelection1.svg

Thales's theorem and reflections

Let Template:Math be a triangle in a circle where Template:Mvar is a diameter in that circle. Then construct a new triangle Template:Math by mirroring Template:Math over the line Template:Mvar and then mirroring it again over the line perpendicular to Template:Mvar which goes through the center of the circle. Since lines Template:Mvar and Template:Mvar are parallel, likewise for Template:Mvar and Template:Mvar, the quadrilateral Template:Mvar is a parallelogram. Since lines Template:Mvar and Template:Mvar, 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.

Fourth proofEdit

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

<math>\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}\qquad \qquad \overrightarrow{CB} = \overrightarrow{CO} + \overrightarrow{OB}</math>

and their dot product can be expanded as

<math>\overrightarrow{AB}\cdot\overrightarrow{CB} = \left(\overrightarrow{AO} + \overrightarrow{OB}\right)\cdot\left( \overrightarrow{CO} + \overrightarrow{OB}\right) =\overrightarrow{AO}\cdot\overrightarrow{CO}+(\overrightarrow{AO}+\overrightarrow{CO})\cdot \overrightarrow{OB}+\overrightarrow{OB}\cdot\overrightarrow{OB}</math>

but

<math>\overrightarrow{AO}=-\overrightarrow{CO}\qquad\qquad \overrightarrow{AO}\cdot\overrightarrow{CO}=-R^2\qquad\qquad \overrightarrow{OB}\cdot\overrightarrow{OB} = R^2</math>

and the dot product vanishes

<math>\overrightarrow{AB}\cdot\overrightarrow{CB} = -R^2+\overrightarrow{0}\cdot \overrightarrow{OB}+R^2=0</math>

and then the vectors <math>\overrightarrow{AB}</math> and <math>\overrightarrow{CB}</math> are orthogonal and the angle ABC is a right angle.

ConverseEdit

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. Template:Collapse top 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. Template: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.)

Proof of the converse using geometryEdit

File:Thales' Theorem Converse.svg

Figure for the proof of the converse

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 Template:Math, Template:Mvar a line parallel to Template:Mvar passing by Template:Mvar, and Template:Mvar a line parallel to Template:Mvar passing by Template:Mvar. Let Template:Mvar be the point of intersection of lines Template:Mvar and Template:Mvar. (It has not been proven that Template:Mvar lies on the circle.)

The quadrilateral Template:Mvar forms a parallelogram by construction (as opposite sides are parallel). Since in a parallelogram adjacent angles are supplementary (add to 180°) and Template:Math is a right angle (90°) then angles Template:Math are also right (90°); consequently Template:Mvar is a rectangle.

Let Template:Mvar be the point of intersection of the diagonals Template:Mvar and Template:Overline. Then the point Template:Mvar, by the second fact above, is equidistant from Template:Mvar, Template:Mvar, and Template:Mvar. And so Template:Mvar is center of the circumscribing circle, and the hypotenuse of the triangle (Template:Mvar) is a diameter of the circle.

Alternate proof of the converse using geometryEdit

Given a right triangle Template:Mvar with hypotenuse Template:Mvar, construct a circle Template:Math whose diameter is Template:Mvar. Let Template:Mvar be the center of Template:Math. Let Template:Mvar be the intersection of Template:Math and the ray Template:Mvar. By Thales's theorem, Template:Math is right. But then Template:Mvar must equal Template:Mvar. (If Template:Mvar lies inside Template:Math, Template:Math would be obtuse, and if Template:Mvar lies outside Template:Math, Template:Math would be acute.)

Proof of the converse using linear algebraEdit

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 Template:Math and circle Template:Mvar with Template:Overline as a diameter. Let M's center lie on the origin, for easier calculation. Then we know

Template:Math, because the circle centered at the origin has Template:Mvar as diameter, and
Template:Math, because Template:Math is a right angle.

It follows

<math>\begin{align}

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

 &= (A-B) \cdot (B+A) \\
 &= |A|^2 - |B|^2. \\[4pt]
 \therefore \ |A| &= |B|.

\end{align}</math>

This means that Template:Mvar and Template:Mvar are equidistant from the origin, i.e. from the center of Template:Mvar. Since Template:Mvar lies on Template:Mvar, so does Template:Mvar, and the circle Template:Mvar 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.

Generalizations and related resultsEdit

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 Template:Mvar, Template:Mvar and Template:Mvar on a circle with center Template:Mvar, the angle Template:Math is twice as large as the angle Template:Math.

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

If Template:Mvar is a diameter of a circle, then:

If Template:Mvar is inside the circle, then Template:Math
If Template:Mvar is on the circle, then Template:Math
If Template:Mvar is outside the circle, then Template:Math.

ApplicationsEdit

Constructing a tangent to a circle passing through a pointEdit

File:Thales' Theorem Tangents.svg

Constructing a tangent using Thales's theorem.

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 Template:Mvar with centre Template:Mvar and the point Template:Mvar outside Template:Mvar, bisect Template:Mvar at Template:Mvar and draw the circle of radius Template:Mvar with centre Template:Mvar. Template:Mvar is a diameter of this circle, so the triangles connecting OP to the points Template:Mvar and Template:Mvar where the circles intersect are both right triangles.

File:Root construction geometric mean5.svg

Geometric method to find <math>\sqrt{p}</math> using the geometric mean theorem <math>h=\sqrt{pq}</math> with <math>q=1</math>

Finding the centre of a circleEdit

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.<ref>Resources for Teaching Mathematics: 14–16 Colin Foster</ref> 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

Illustration of the use of Thales's theorem and a right angle to find the centre of a circle