Thales's theorem

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

${\displaystyle \alpha +\left(\alpha +\beta \right)+\beta =180^{\circ }}$

${\displaystyle 2\alpha +2\beta =180^{\circ }}$

${\displaystyle 2(\alpha +\beta )=180^{\circ }}$

${\displaystyle \therefore \alpha +\beta =90^{\circ }.}$

Q.E.D.

The theorem may also be proven using trigonometry: Let ${\displaystyle O=(0,0)}$, ${\displaystyle A=(-1,0)}$, and ${\displaystyle C=(1,0)}$. Then B is a point on the unit circle ${\displaystyle (\cos \theta ,\sin \theta )}$. 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:

${\displaystyle m_{AB}={\frac {y_{B}-y_{A}}{x_{B}-x_{A}}}={\frac {\sin \theta }{\cos \theta +1}}}$

and

${\displaystyle m_{BC}={\frac {y_{B}-y_{C}}{x_{B}-x_{C}}}={\frac {\sin \theta }{\cos \theta -1}}}$

Then we show that their product equals −1:

${\displaystyle {\begin{aligned}&m_{AB}\cdot m_{BC}\\[8pt]={}&{\frac {\sin \theta }{\cos \theta +1}}\cdot {\frac {\sin \theta }{\cos \theta -1}}\\[8pt]={}&{\frac {\sin ^{2}\theta }{\cos ^{2}\theta -1}}\\[8pt]={}&{\frac {\sin ^{2}\theta }{-\sin ^{2}\theta }}\\[8pt]={}&{-1}\end{aligned}}}$

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

Thales's theorem and reflections

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

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 ∠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 (Note that 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, and ∠ADC are also right (90°); consequently ABCD is a rectangle.

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.

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.)

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

0 = (A − B) · (B − C) = (A − B) · (B + A) = |A|² − |B|².

Hence:

|A| = |B|.

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.