Editing Thales's theorem (section)

== Proof ==

===First proof===
The following facts are used: the [[sum of angles of a triangle|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 angle|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

:<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 proof===
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 [[slope]]s is equal to −1. We calculate the slopes for {{mvar|{{overline|AB}}}} and {{mvar|{{overline|BC}}}}:

:<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 proof===
[[File:Thales_theorem_by_refelection1.svg|thumb|Thales's theorem and reflections]]
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 (geometry)|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.

===Fourth proof===
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.