Difference between revisions of "Thales' theorem"

Latest revision as of 23:42, 4 June 2021

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 proven by considering that the intercepted arc is a semicircle and has measure $180^{\circ}$ . Thus, the intercepted angle has degree measure $\frac{180}{2} = 90$ .

This theorem has many uses in geometry because it helps introduce right angles into problems; however, the name of the theorem is not well-known. Thus, you may cite the "universal fact" that $\angle ABC = 90^{\circ}$ in proofs without specifically referring to Thales.

Problems

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! Thales

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-Thale's Theorem states that if there are three points on a circle, <math>A,B,C</math> with <math>AC</math> being a diameter, <math>\angle ABC=90^{\circ}</math>
+[[Thales]]' Theorem states that if there are three points on a circle, <math>A, B, C</math> with <math>AC</math> being a diameter, <math>\angle ABC=90^{\circ}</math>.
@@ Line 17: / Line 17: @@
-{{stub}}
+This is proven by considering that the intercepted arc is a semicircle and has measure <math>180^{\circ}</math>. Thus, the intercepted angle has degree measure <math>\frac{180}{2} = 90</math>.
+This theorem has many uses in geometry because it helps introduce right angles into problems; however, the name of the theorem is not well-known. Thus, you may cite the "universal fact" that <math>\angle ABC = 90^{\circ}</math> in proofs without specifically referring to Thales.
+==Problems==
+. Prove that the converse of the theorem holds: if <math>\angle ABC = 90^{\circ}</math>, <math>AC</math> is a diameter.
+. Prove that if rectangle <math>ABCD</math> is inscribed in a circle, then <math>AC</math> and <math>BD</math> are diameters. (Thus, <math>AC = BD</math>.)
+. <math>AC</math> is a diameter to circle O with radius 5. If B is on O and <math>AB = 6</math>, then find <math>BC</math>.
+. Prove that in a right triangle with AD the median to the hypotenuse, <math>AD = BD = CD</math>.
+. <math>AC</math> is a diameter to circle O, B is on O, and D is on the extension of segment <math>BC</math> such that <math>AD</math> is tangent to O. If the radius of O is 5 and <math>AD = 24</math>, find <math>AB</math>.
+. In a triangle <math>ABC</math>, <math>CD</math> is the median to the side <math>AB</math>(<math>D</math> is the midpoint). If <math>CD=BD</math>, then prove that <math>\angle C=90^\circ</math> 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!''
+[[Thales]]

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Difference between revisions of "Thales' theorem"

Latest revision as of 23:42, 4 June 2021

Problems