Thales's theorem



In geometry, Thales's theorem states that if $\overline{AC}$, $B$, and $A$ are distinct points on a circle where the line $B$ is a diameter, the angle $∠ ABC$ 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. It is generally attributed to Thales of Miletus, but it is sometimes attributed to Pythagoras.

History


Babylonian mathematicians knew this for special cases before Greek mathematicians proved it.

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. 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".

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.

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

First proof
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.

Since $∠ ABC$, $\overline{OA} = \overline{OB} = \overline{OC}$ and $△OBA$ are isosceles triangles, and by the equality of the base angles of an isosceles triangle, $△OBC$ and $∠ OBC = ∠ OCB$.

Let $∠ OBA = ∠ OAB$ and $α = ∠ BAO$. The three internal angles of the $β = ∠ OBC$ triangle are $C$, $∆ABC$, and $\overline{AC}$. Since the sum of the angles of a triangle is equal to 180°, we have


 * $$\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.

Second proof
The theorem may also be proven using trigonometry: Let $(α + β)$, $O = (0, 0)$, and $A = (-1, 0)$. Then $\overline{AC}$ is a point on the unit circle $C = (1, 0)$. We will show that $(cos θ, sin θ)$ forms a right angle by proving that $B$ and $α$ are perpendicular — that is, the product of their slopes is equal to −1. We calculate the slopes for $β$ and $B$:


 * $$\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:


 * $$\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 $$\sin^2 \theta + \cos^2 \theta = 1.$$

Third proof
Let $△ABC$ be a triangle in a circle where $\overline{AB}$ is a diameter in that circle. Then construct a new triangle $△ABC$ by mirroring $△ABD$ over the line $\overline{BC}$ and then mirroring it again over the line perpendicular to $\overline{AB}$ which goes through the center of the circle. Since lines $\overline{BC}$ and $AB$ are parallel, likewise for $AB$ and $AB$, the quadrilateral $AC$ is a parallelogram. Since lines $BD$ and $AD$, 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.

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

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 geometry


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 $△ABC$, $CB$ a line parallel to $ACBD$ passing by $AB$, and $CD$ a line parallel to $r$ passing by $\overline{BC}$. Let $A$ be the point of intersection of lines $s$ and $\overline{AB}$. (It has not been proven that $C$ lies on the circle.)

The quadrilateral $D$ forms a parallelogram by construction (as opposite sides are parallel). Since in a parallelogram adjacent angles are supplementary (add to 180°) and $∠ ABC$ is a right angle (90°) then angles $∠ ABC$ are also right (90°); consequently $r$ is a rectangle.

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

Alternate proof of the converse using geometry
Given a right triangle $B$ with hypotenuse $C$, construct a circle $∠ BAD, ∠ BCD, ∠ ADC$ whose diameter is $O$. Let $\overline{AC}$ be the center of $Ω$. Let $ABC$ be the intersection of $Ω$ and the ray $AC$. By Thales's theorem, $Ω$ is right. But then $AC$ must equal $O$. (If $D$ lies inside $∠ ADC$, $△ABC$ would be obtuse, and if $OB$ lies outside $∠ ADC$, $△ABC$ would be acute.)

Proof of the converse using linear algebra
This proof utilizes two facts: Let there be a right angle $∠ ADC$ and circle $D$ with $B$ as a diameter. Let M's center lie on the origin, for easier calculation. Then we know It follows
 * 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.
 * $∠ ABC$, because the circle centered at the origin has $D$ as diameter, and
 * $A = −C$, because $(A – B) · (B – C) = 0$ is a right angle.
 * $$\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 $D$ and $M$ are equidistant from the origin, i.e. from the center of $\overline{AC|}$. Since $\overline{AC}$ lies on $A$, so does $B$, and the circle $M$ 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.

Generalizations and related results
Thales's theorem is a special case of the following theorem:
 * Given three points $A$, $M$ and $B$ on a circle with center $M$, the angle $∠ ABC$ is twice as large as the angle $∠ AOC$.

See inscribed angle, the proof of this theorem is quite similar to the proof of Thales's theorem given above.

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


 * If $A$ is a diameter of a circle, then:
 * If $B$ is inside the circle, then $∠ ABC$
 * If $C$ is on the circle, then $∠ ABC > 90°$
 * If $O$ is outside the circle, then $∠ ABC = 90°$.

Constructing a tangent to a circle passing through a point
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 $\overline{AC}$ with centre $B$ and the point $B$ outside $B$, bisect $k$ at $O$ and draw the circle of radius $P$ with centre $k$. $\overline{OP}$ is a diameter of this circle, so the triangles connecting OP to the points $H$ and $\overline{OH}$ where the circles intersect are both right triangles.



Finding the centre of a circle
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. 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.