Thales' theorem

Revision as of 23:17, 29 November 2016 by Gimbob (talk | contribs) (added one more problem)

Thales' Theorem states that if there are three points on a circle, $A,B,C$ with $AC$ being a diameter, $\angle ABC=90^{\circ}$.

[asy] dot((5,0)); dot((-5,0)); draw(circle((0,0),5)); dot((3,4)); draw((5,0)--(3,4)); draw((-5,0)--(3,4)); draw((-5,0)--(5,0)); label("A",(-5,0),W); label("B",(3,4),NE); label("C",(5,0),E); [/asy]

This is easily proven by considering that the intercepted arc is a semicircle, or 180°. Thus, the intercepted angle is 180°/2 = 90°.

This theorem has many uses in geometry because it helps introduce right angles into a problems; however, the name of the theorem is not well-known. Thus, you may cite the "universal fact" that <ABC = 90° in a proof without specifically referring to Thales.


1. Prove that the converse of the theorem holds: if $\angle ABC = 90^{\circ}$, $AC$ is a diameter.

2. Prove that if rectangle $ABCD$ is inscribed in a circle, then $AC$ and $BD$ are diameters. (Thus, $AC = BD$.)

3. $AC$ is a diameter to circle O with radius 5. If B is on O and $AB = 6$, then find $BC$.

4. Prove that in a right triangle with AD the median to the hypotenuse, $AD = BD = CD$.

5. $AC$ is a diameter to circle O, B is on O, and D is on the extension of segment $BC$ such that $AD$ is tangent to O. If the radius of O is 5 and $AD = 24$, find $AB$.

6. In a triangle $ABC$, $CD$ is the median to the side $AB$($D$ is the midpoint). If $CD=BD$, then prove that $\angle C=90^\circ$ without using Thales' theorem. If you have a general understanding of how the theorem works and its proof you can manipulate it into the solution.

Please add more problems!

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