Thales' Theorem

[Image Credits: Wikipedia](https://en.wikipedia.org/wiki/Thales%27s<em>theorem#/media/File:Animated</em>illustration<em>of</em>thales_theorem.gif) Image Credits: Wikipedia

Thales' Theorem is a simple yet important geometrical result which describes the relationship between inscribed right angles and circles. A special case of the inscribed angle theorem, Thales' Theorem is as follows:

Let AA and CC be two points on the circumference of some circle SS such that AC\overline{AC} is a diameter of SS. If a point BB is placed anywhere on the circumference of SS (other than AA or CC), then ABC\angle ABC will always be a right angle.

Using the inscribed angle theorem, it also follows that the reverse of this statement holds. Given any 3 points A,B,CA,B,C on the circumference of SS such that ABC=90\angle ABC=90^{\circ}, AC\overline{AC} must be a diameter of the circle. We now present two different approaches to proving Thales'.

A proof via inscribed angle theorem.

Consider a circle SS with center OO and two diametrically opposite points A,CA,C on the circumference ofSS. Since points A,C,OA,C,O are all co-linear, it follows that AOC=180\angle AOC=180^{\circ}.

By the inscribed angle theorem, we know that any angle θ\theta inscribed in SS will be half of the central angle which subtends the same arc on SS. In other words, if a point BB is placed on the circumference of SS, then angle ABC\angle ABC will be half of angle AOC\angle AOC. Since we have that AOC=180\angle AOC=180^{\circ}, it follows that ABC=90\angle ABC=90^{\circ}

Thales' theorem has many elegant applications to pure geometry. However, since Thales' often arises in coordinate geometry, let's take a look at a proof of Thales' which uses just that.

A proof via coordinate geometry.

Consider a circle SS of radius rr located at the origin. Notice that SS is described by the equation x2+y2=r2x^2+y^2=r^2. Consider the two diametrically opposite points (r,0)(-r,0) and (r,0)(r,0) located at the intersection of SS with the xx-axis. Place a third point anywhere on the upper half of the circumference of SS.

Pythagoras' theorem guarantees that in a triangle with side lengths a,b,ca,b,c, with cc being the longest side, a2+b2=c2a^2+b^2=c^2 if and only if the triangle is a right triangle. Therefore, we will show that θ=90\theta=90^{\circ} in the diagram to the right by showing that a2+b2=c2a^2+b^2=c^2.

Using Pythagoras' theorem, we can determine the legnths of each side as follows

a2=(x(r))2+(y0)2=x2+2xr+r2+y2b2=(xr)2+(y0)2=x22xr+r2+y2c2=(r(r))2=4r2\begin{aligned} a^2&=(x-(-r))^2+(y-0)^2=x^2+2xr+r^2+y^2\\b^2&=(x-r)^2+(y-0)^2=x^2-2xr+r^2+y^2\\c^2&=(r-(-r))^2=4r^2\end{aligned}

Next, observe that

a2+b2=(x2+2xr+r2+y2)+(x22xr+r2+y2)=2x2+2y2+2r2=2(x2+y2)+2r2a^2+b^2=(x^2+2xr+r^2+y^2)+(x^2-2xr+r^2+y^2)=2x^2+2y^2+2r^2=2(x^2+y^2)+2r^2

Recall that since the point (x,y)(x,y) was chosen on the circumference of SS, it must satisfy x2+y2=r2x^2+y^2=r^2. Using this to simplify the equation above:

a2+b2=2(x2+y2)+2r2=2(r2)+2r2=4r2=c2a^2+b^2=2(x^2+y^2)+2r^2=2(r^2)+2r^2=4r^2=c^2

Since we have shown that a2+b2=c2a^2+b^2=c^2, we have also proved that θ=90\theta=90^{\circ} which finishes our proof of Thales'.

There are many other ways to prove Thales', some using pure geometry, some using trigonometry, and some even using complex numbers! We encourage you to explore and try to come up with your own unique proof of Thales'. If you do manage to come up with your own proof, great! Just don't get too excited and sacrifice an ox (see entry "Thales' theorems" on this page for more details).

Here are a couple problems centered around Thales' for practice. The first few aren't too bad, but be careful, the last one is quite tricky!

Circle SS is placed in the 2D-plane with radius 33 and its center OO placed on the yy-axis. SS intersects the yy-axis at both the point GG and the origin (labeled FF).

Circle TT is then drawn with twice the radius of circle SS such that TT also intersects FF and GG.

If circle TT intersects the xx-axis at points XX and GG, find the length of segment FX\overline{FX}.



Circles SS and TT are co-centric at point OO with circle SS inside circle TT. Line segment BA\overline{BA} is a cord of circle TT and is tangent to circle SS at point DD. Line segment BC\overline{BC} is a cord of circle TT and is perpendicular to segment BABA.

If BC\overline{BC} has length 1 and BA\overline{BA} has length 2, find the radius of the inner circle SS




This last one can be a bit tricky.

Circle TT is drawn with perpendicular cords AB\overline{AB} and BC\overline{BC} with lengths 3 and 4, respectively. Circle SS is then drawn such that AB\overline{AB} is a diameter of SS. Finally, a line is drawn through point AA perpendicular to AC\overline{AC}. Let point EE be the other intersection of this line with circle SS.

Find the length of segment BE\overline{BE}.




Note by Trevor Arashiro
2 months, 2 weeks ago

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For the last question, there is a slicker solution involving similar triangles (ABCBEA\triangle ABC \sim \triangle BEA), and the answer immediately follows after finding the length ACAC.

Elijah L - 2 months, 1 week ago

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There certainly is! However, i wrote this article assuming nothing about readers’ background in geometry. Many people learn Thales’ before similar triangles so I wanted to make sure everyone could understand it.

Trevor Arashiro - 2 months, 1 week ago

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@Trevor Arashiro How you made these switches of 'solution' and 'hint'?

Zakir Husain - 1 month, 3 weeks ago

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Yes, even I want to know!

Mahdi Raza - 1 month, 3 weeks ago

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The last question has a solution using trigonometry as well, and the answer comes out without much effort

Sachetan Debray - 1 month, 2 weeks ago

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But I agree that your solution is much more elegant and also useful for learners of pure geometry

Sachetan Debray - 1 month, 2 weeks ago

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