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Mathematics

Mathematics

Thales' theorem

Every point on a semicircle sees the diameter at 90°: the simplest way to get an exact right angle.

Draw a circle, run a diameter from AA to BB and put a point CC anywhere on the upper arc. Join CC to AA and to BB and a triangle appears. Now measure the angle at CC: it is always exactly 90°, no matter where on the arc you placed CC. This discovery is credited to Thales of Miletus, it is about 2600 years old and still the fastest route to a clean right angle.

What the theorem actually says

More precisely: if the segment ABAB is a diameter and CC lies on the circle but neither on AA nor on BB, then the triangle ABCABC is right-angled, and the right angle sits at CC. The diameter ABAB is therefore automatically the longest side, the . The two shorter sides ACAC and BCBC are called the legs. In this role the circle is called the Thales circle. Both conditions matter: ABAB really has to pass through the centre, and CC has to sit exactly on the arc. So in a circle of diameter 10 cm every such triangle has a of 10 cm, only the two legs change depending on where CC sits.

ACB=90\angle ACB = 90^\circ
If CC lies on the semicircle over the diameter ABAB, the angle at CC is a right angle.
Circle gauge
Radius r4
r = 4
U = 2 · π · r = 25.1A = π · r² = 50.3
Try it: set r=5r = 5. The radius drawn in the picture runs from the centre to the rim, and wherever on the rim you turn it, it stays 5 long. The diameter is twice as long, so 10. That always equal distance to the centre is exactly what creates the right angle.

Why it always works

Call the centre MM. Because AA, BB and CC all lie on the circle, the segments MAMA, MBMB and MCMC are equally long, they are all radii. So the triangle AMCAMC is isosceles, which means the angles at AA and at CC are equal, call them α\alpha. In the same way BMCBMC is isosceles, and there the two equal angles are called β\beta. In the big triangle ABCABC the angle at AA is α\alpha, the angle at BB is β\beta, and at CC the two meet, so there it is α+β\alpha + \beta. Since all three angles together make 180°, only half of that is left for α+β\alpha + \beta.

2α+2β=180α+β=902\alpha + 2\beta = 180^\circ \quad\Rightarrow\quad \alpha + \beta = 90^\circ
The angle at CC is α+β\alpha + \beta, and that is pinned down: 90°.

Building right angles, finding centres

Above all the theorem is a tool. First, it lets you construct right angles: draw a segment, halve it, swing a circle around the midpoint through both end points, and every point of the arc hands you an exact right angle, with no set square at all. Second, the converse holds: if a triangle has a right angle at CC, then CC lies on the circle over the hypotenuse. The midpoint of the hypotenuse is therefore the same distance from all three corners, namely half the hypotenuse. For a hypotenuse of 13 cm that is 6.5 cm. If you only know the two legs aa and bb instead of the hypotenuse, you get the hypotenuse cc first from Pythagoras' theorem, a2+b2=c2a^2 + b^2 = c^2. Third, this is how you find the centre of a round table top: place the right angle of a set square with its corner on the rim, mark the two spots where its arms cross the rim, and join them. That line is a diameter. Repeat it somewhere else, and the two diameters cross exactly at the centre.

Exercises

0 of 6 solved

Time to try it yourself. You can't break anything, every attempt counts.

A triangle has the diameter of a circle as one side, its third corner lies on the circle. Where is the right angle?

A Thales circle has radius 6 cm. How long is the hypotenuse of every triangle you draw in it over the diameter, in cm?

In a Thales triangle over the diameter ABAB the angle at AA is 35°. How large is the angle at BB in degrees?

In a Thales triangle the diameter ABAB is always the .

You put CC not on the rim but somewhere inside the circle, though not on the segment ABAB, and join it to AA and BB. How large is the angle at CC?

Match each part of the Thales triangle to what it is.

Diameter ABAB
Point CC on the arc
Centre MM
Segment MCMC