Editing Thales's theorem (section)

===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.