Thales's theorem

Proof

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 OA = OB = OC, △OBA and △OBC are isosceles triangles, and by the equality of the base angles of an isosceles triangle, ∠ OBC = ∠ OCB and ∠ OBA = ∠ OAB.

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

${\displaystyle \begin{array}{cc}\alpha +(\alpha +\beta )+\beta & ={180}^{\circ }\\ 2\alpha +2\beta & ={180}^{\circ }\\ 2(\alpha +\beta )& ={180}^{\circ }\\ \therefore \alpha +\beta & ={90}^{\circ }.\end{array}}$

Q.E.D.

Second proof

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

${\displaystyle \begin{array}{cc}{m}_{AB}& =\frac{y_B-y_A}{x_B-x_A}=\frac{\sin \theta }{\cos \theta +1}\\ m_{BC}& =\frac{y_C-y_B}{x_C-x_B}=\frac{-\sin \theta }{-\cos \theta +1}\end{array}}$

Then we show that their product equals −1:

${\displaystyle \begin{array}{cc}& m_{AB}\cdot m_{BC}\\ =& \frac{\sin \theta }{\cos \theta +1}\cdot \frac{-\sin \theta }{-\cos \theta +1}\\ =& \frac{-{\sin }^2\theta }{-{\cos }^2\theta +1}\\ =& \frac{-{\sin }^2\theta }{{\sin }^2\theta }\\ =& -1\end{array}}$

Note the use of the Pythagorean trigonometric identity ${\displaystyle {\sin }^2\theta +{\cos }^2\theta =1.}$

Third proof

Let △ABC be a triangle in a circle where AB is a diameter in that circle. Then construct a new triangle △ABD by rotating △ABC by 180° over the center of the circle O. Since we rotated over 180°, lines AC and BD are parallel, likewise for AD and CB. It follows that the quadrilateral ACBD is a parallelogram. The lines OC and OD are likewise parallel and meet in O, so CD is a straight line through O, meaning CD is a diameter of the circle. Since lines AB and 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.

Fourth proof

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

${\displaystyle \overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}\overrightarrow{CB}=\overrightarrow{CO}+\overrightarrow{OB}}$

and their dot product can be expanded as

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

but

${\displaystyle \overrightarrow{AO}=-\overrightarrow{CO}\overrightarrow{AO}\cdot \overrightarrow{CO}=-R^2\overrightarrow{OB}\cdot \overrightarrow{OB}=R^2}$

and the dot product vanishes

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

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

Converse

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. Its center is called the circumcenter, which is the intersection point of the perpendicular bisectors of the triangle.

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

The converse of Thales's theorem is then: the circumcenter 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, r a line parallel to BC passing by A, and s a line parallel to AB passing by C. Let D be the point of intersection of lines r and s. (It has not been proven that D lies on the circle.)

The quadrilateral ABCD 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 ∠ BAD, ∠ BCD, ∠ ADC are also right (90°); consequently ABCD is a rectangle and D lies on the circle.

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

Alternate proof of the converse using geometry

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

Proof of the converse using linear algebra

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 ∠ ABC and circle M with AC as a diameter. Let M's center lie on the origin, for easier calculation. Then we know

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

It follows

${\displaystyle \begin{array}{cc}0& =(A-B)\cdot (B-C)\\ & =(A-B)\cdot (B+A)\\ & =|A{|}^2-|B{|}^2.\\ \therefore |A|& =|B|.\end{array}}$

This means that A and B are equidistant from the origin, i.e. from the center of M. Since A lies on M, 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

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 A, B and C on a circle with center O, the angle ∠ AOC is twice as large as the angle ∠ ABC.

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

• If AC is a diameter of a circle, then:

• If B is inside the circle, then ∠ ABC > 90°
• If B is on the circle, then ∠ ABC = 90°
• If B is outside the circle, then ∠ ABC < 90°.

Applications

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 k with centre O and the point P outside k, bisect OP at H and draw the circle of radius OH with centre H. OP is a diameter of this circle, so the triangles connecting OP to the points T and T′ 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.