# Exploring Thales’ Theorem

I was playing with a geometry software package and decided to explore Thales Theorem.

The theorem states that for any diameter line drawn through the circle with endpoints B and C on the circle (obviously passing through the circle’s center point), any third non-collinear point A on the circle can be used to form a right angle triangle. That is, no matter where you place A on the circle, the angle BAC is always a right angle. Most places I have read online stop there. There was one small problem on my software. Since constructing this circle meant that the center point was already defined on my program, there didn’t seem to be a way to make the center point part of the line, except by manipulating the mouse or arrow keys. So, as a result, my angle ended up being slightly off: $90.00550^{\circ}$ was the best I could do. But then, I noticed something else: No matter where point A was moved from then on, the angle would stay exactly the same, at $90.00550^{\circ}$. Now, $90.00550^{\circ}$ is not a right angle. Right angles have to be exactly $90^{\circ}$ or go home. If it’s not a right angle, then Thales’ theorem should work for any angle.

Why not restate the theorem for internal angles in the circle a little more generally then?

For any chord with endpoints BC in the circle, and a point A in the major arc of the circle, all angles $\angle BAC$ will all equal some angle $\theta$. For points A in the minor arc, all angles will be equal to $180^{\circ} - \theta$. Note that BC is the desired chord, making the arc containing point A the major arc, with the small arc in the lower part of the circle the minor arc. As shown, all angles in the major arc are about 30.3 degrees.

So, now the limitations of my software are unimportant. In the setup shown on the left, the circle contains the chord BC, and A lies in the major arc, forming an angle $\angle BAC = 30.29879^{\circ}$. If A lay in the minor arc, the angle would have been $180^{\circ} - 30.29879^{\circ} = 149.70121^{\circ}$.

By manipulating BC, you can obtain any angle $\angle BAC$ you like, so long as $\angle BAC < 180^{\circ}$. More precisely, all angles in the minor arc drawn in the manner previously described will be $90^{\circ} < \angle BAC < 180^{\circ}$, and all angles in the major arc will tend to be: $0^{\circ} < \angle BAC < 90^{\circ}$. If the chord is actually the diameter line of the circle, then $\angle BAC = 90^{\circ}$ exactly.

# Doing math without “technology”

Math is a mental process. Basic math techniques should be, for those that are learning basic math, of the variety that you should either be able to do in your head, writing your thoughts on paper; or learn how to do so if you can’t yet. It is essentially a human skill. We acquire new math skills by building upon a foundation of what we have previously learned. In my opinion, there is nothing that modern computers can do to change this, although it might help us learn different things.

Math is a means of pure intellectual inquiry. How would you investigate the following:

A is 100 m to the north of B. A moves to the east at 3 m/s, while B moves to the west at 3 m/s. Show that the midpoint of the line AB is fixed under these conditions regardless of the position of A or B.

There are a number of ways to show this to yourself. I didn’t want to say “prove”, since I am avoiding a rigorous proof here. Just say that one rainy day, you simply asked yourself that question. You weren’t trying to be rigorous, just curious. There is no teacher in the room; no critical parents. Just you, a pencil and some paper. And maybe a ruler. And an eraser. How would you tackle it?

Some people would draw a diagram moving A and B apart in regular invervals, with $\overline{AB}$ lines drawn between them. Others would show that the areas of the two triangles described by the initial vertical line $\overline{AB}$ at time zero and their present position are the same area, which implies their hypoteneuses are equal. You can also show that the midpoint of the line $\overline{AB}$ at any time > 0 is the same midpoint as exists between the initial $\overline{AB}$ at time = 0.