Editing Thales's theorem (section)

==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.
{{collapse top|title=Uniqueness proof (sketch)}}
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.
{{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 geometry===
[[Image:Thales' Theorem Converse.svg|thumb|200px|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 {{math|∠ ''ABC''}}, {{mvar|r}} a line parallel to {{mvar|{{overline|BC}}}} passing by {{mvar|A}}, and {{mvar|s}} a line parallel to {{mvar|{{overline|AB}}}} passing by {{mvar|C}}. Let {{mvar|D}} be the point of intersection of lines {{mvar|r}} and {{mvar|s}}. (It has not been proven that {{mvar|D}} lies on the circle.)

The quadrilateral {{mvar|ABCD}} forms a parallelogram by construction (as opposite sides are parallel). Since in a parallelogram adjacent angles are supplementary (add to 180°) and {{math|∠ ''ABC''}} is a right angle (90°) then angles {{math|∠ ''BAD'', ∠ ''BCD'', ∠ ''ADC''}} are also right (90°); consequently {{mvar|ABCD}} is a rectangle.

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

===Alternate proof of the converse using geometry===
Given a right triangle {{mvar|ABC}} with hypotenuse {{mvar|AC}}, construct a circle {{math|Ω}} whose diameter is {{mvar|AC}}. Let {{mvar|O}} be the center of {{math|Ω}}. Let {{mvar|D}} be the intersection of {{math|Ω}} and the ray {{mvar|OB}}. By Thales's theorem, {{math|∠ ''ADC''}} is right. But then {{mvar|D}} must equal {{mvar|B}}. (If {{mvar|D}} lies inside {{math|△''ABC''}}, {{math|∠ ''ADC''}} would be obtuse, and if {{mvar|D}} lies outside {{math|△''ABC''}}, {{math|∠ ''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 [[vector (geometry)|vector]]s 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 {{math|∠ ''ABC''}} and circle {{mvar|M}} with {{overline|AC|}} as a diameter.
Let M's center lie on the origin, for easier calculation.
Then we know
*{{math|1=''A'' = −''C''}}, because the circle centered at the origin has {{mvar|{{overline|AC}}}} as diameter, and
*{{math|1=(''A'' – ''B'') · (''B'' – ''C'') = 0}}, because {{math|∠ ''ABC''}} 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 {{mvar|A}} and {{mvar|B}} are equidistant from the origin, i.e. from the center of {{mvar|M}}. Since {{mvar|A}} lies on {{mvar|M}}, so does {{mvar|B}}, and the circle {{mvar|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]].