## Theorem 32: the angle at the centre is twice the angle at the circumference

I recently watched someone do a session with some 15 and 16 year olds in which she used the dynamic geometry software GeoGebra.  I’ve seen dynamic software before (more about that in a moment), but the great thing about this particular package is that it’s free, and very quick and easy to download.   It’s also very easy to use: I produced the diagrams for this post very quickly, without having to read any help files or instructions!

I think the first time I used any software like this was at a Royal Institution masterclass, when I was about 14 (which was, err, cough, quite a long time ago).  I remember quite a lot about this masterclass, despite it happening more than ten years ago.  I don’t remember all of what we did, but I do remember that we were encouraged to use the software to explore interesting properties of circles and angles.  (I’m not absolutely sure what software we used.  It might have been some version of Cabri.)  One of the nice things about this software (and GeoGebra) is that one can measure angles and lengths, and then move points (or other objects) around and see how the angles and lengths change.  Here’s an example (that we looked at during the masterclass).  Unfortunately, the screenshots that I can include here don’t give you the opportunity to try moving things for yourself, but you can easily download GeoGebra and do it that way. The point $O$ is the centre of the circle, and $A$, $B$ and $C$ lie on the circumference.  I’ve chosen a couple of angles to measure.  (We have to remember that the computer cannot measure an angle absolutely exactly, but these measurements should be pretty close.)

Let’s see what happens if I move point $C$ around on the circumference.  Here are a few screenshots for you (in case you don’t want to try it for yourself).  My recollection of this masterclass is that we were encouraged to see what we noticed, and then try to explain and prove anything relevant.  If you have GeoGebra (or similar), you could try moving $A$ and $B$ too, to see what happens.

So I suggest you try this for yourself!  More details over the fold.

In all but one of those pictures, the angle $\angle ACB$ seemed to be very close to being half of $\angle AOB$, didn’t it?  (This might be more clear if you tried moving $A$ and $B$ too, to check that it wasn’t a particular feature of the points that I used.)  The exception was the last diagram.  Perhaps it would help if I measured slightly different angles: Ah!

So I think we now have a conjecture, which we could express as the angle at the centre is twice the angle at the circumference.  Can we prove it?  Can you prove it?!

I think this is why I remember the masterclass, because I think I did just about get a proof before we ran out of time, and I think it was possibly the first time I’d come up with a proof.  The idea that I could be completely sure that this would always work, for infinitely many points on the circle (not just the ones I’d tried on the computer), was very cool!  I still think that, in fact.  Which is why I’m writing this post.

So, how can we go about proving this?  I’ve drawn in the radius $OC$ in this next diagram, because I can see that it will split the quadrilateral into two isosceles triangles, and perhaps those will help us to understand the angles.  I’ve labelled $\angle ACO$ as $\alpha_1$ and $\angle OCB$ as $\beta_1$. Now let’s use those isosceles triangles: they look much too good to waste.   Which triangles?  I’m looking at triangle $AOC$, which has $AO=OC$ (both lengths are radii of the circle), and triangle $COB$, which has $CO=OB$.  So we see that $\angle ACO=\angle CAO$ and $\angle CBO = \angle BCO$.  On the diagram below, $\alpha_1=\alpha_2$ and $\beta_1=\beta_2$.  (I’d have used just $\alpha$ twice and $\beta$ twice, but I haven’t managed to get the software to do that!) Now what?  Well, let’s chase some angles.  We’re interested in $\angle ACB$ and $\angle AOB$.  Using our new Greek letters, we have $\angle ACB = \alpha_1+\beta_1$

The angles at a point add to $360^{\circ}$, so $\angle AOB=360 - \angle AOC - \angle COB$.  But we can find those last two angles, because the angles in each of our isosceles triangles sum to $180^{\circ}$.  One triangle gives $\angle AOC = 180-2\alpha_1$, and the other $\angle COB=180-2\beta_1$.  Substituting these in gives $\angle AOB = 360-(180-2\alpha_1)-(180-2\beta_1) = 2(\alpha_1+\beta_1)$.

That is, $\angle AOB = 2(\alpha_1+\beta_1)=2\angle ACB$, and that’s exactly what we wanted!

Great.  Job done, right?

Not quite.  One important point with geometry problems like this (and I’m sure that I missed this subtlety when I was 14) is that we have to make sure that our arguments work whatever the diagram.

The second diagram I drew above looked rather different to the first: the line segments $CB$ and $AO$ intersected.  So drawing in the radius $OC$ will lead to a rather different diagram from the one we used in the proof above, so we have to check that our proof still works.  It’s not too hard to tweak the argument, but it does need doing.  I’ll leave that for you as a good exercise — it shouldn’t take you long!

And then there’s a third case to consider: the one where I had to redraw the diagram with different angles labelled.  Again, not too hard, so again I’ll leave it for you.

Then we have a theorem.

Theorem The angle at the centre is twice the angle at the circumference.

This is the standard way to express this theorem.  Hopefully it’s now clear to you what it means.

This is a surprisingly useful fact to know (at least if you want to prove theorems about circles).  For example, it immediately allows one to conclude some other interesting things.  I’m not going to give details now, but here’s a diagram that might inspire you: I encourage you to explore.  If you want some hints, you could try Wikipedia.  You could also try introductions to Euclidean geometry for further reading.

### 8 Responses to “Theorem 32: the angle at the centre is twice the angle at the circumference”

1. Olof Says:

I was thinking about this the other day and wanted to justify why one ought to draw the radius OC as the first step of the proof of the theorem. I came to the conclusion that this was a reasonable thing to do because all we know about C is that it is a point on (the circumference of) the circle, and a criterion for whether a point lies on the circle or not is that it should be at a certain distance from the centre (O) of the circle.

In other words, drawing the radius OC is a reasonable step to take because it is an illustration of the given information that C is a point on the circle.

Here’s another (related) problem: if A and B are diametrically opposed points on a circle and C is any other point on the circle, then what can you say about the angle ACB? You might be able to prove the answer in different ways!

2. Theorem 33: the size of Gauss sums « Theorem of the week Says:

[…] Theorem of the week Expositions of interesting mathematical results « Theorem 32: the angle at the centre is twice the angle at the circumference […]

3. bikram mathematician Says:

try ext angle= sum of 2 int angles. do it for 2 sides of same triangle and deduce. u r at the edge and u found it.

4. Rayyanu Abdulkarim Says:

Fantastic solution

5. Chand Sapam Says:

I want to know that

Will the angle made between chord and diameter of a circle be half of the angle at the centre made between the radius and diameter?

6. hope Says:

thanks

7. Answer Says:

how did you get the angle = 209. 05, the angle that extends outside the circle

8. theoremoftheweek Says:

I think the angle is 29.05 not 209.05.