There are various applets available online and indeed good Geogebra constructions that seek to demonstrate the various Circle Theorems.

However, I think it is more powerful to construct them from scratch in front of the class using Geogebra, starting from a blank sheet. It requires a degree of familiarity with Geogebra and reliable IT infrastructure (something which my school sadly lacks) but I think it is a more instructive way of walking through them than just moving some points on an existing animation.

So, this blog post, is an attempt to show my progression through the Circle theorems.

## 1, Set up.

First, get the Geogebra page set up, remove the grid and the axes as they are distracting. Also change the menu options so your screen isn’t littered with labels.

## 2, Draw a circle!

Point a point for the **centre** and then draw a nice big **circle** and point a couple of points on the **circumference**, talking through this as you are doing it using the terms in bold.

## 3, Angles subtended by the same arc are equal

You need three points, but best not to use the one which was used to define the circle as this will change the size of the circle. Two of the points define an **arc**. Then draw two **line segments** to **subtend an angle** from that **arc**.

Then measure the angle between the lines and show how it changes:

Maybe add another point to create a second angle **subtended by the same arc.** Ask students to predict what the angle will be before measuring it.

You might want to highlight the arc itself using the arc tool.

## 4, The angle subtended by a semi-circle is 90° OR The triangle in a semi-circle is right-angled

Next investigate what happens as you adjust the arc and look at the **special case** that occurs when the angle is 90°.

## 5, The Angle subtended at the centre is double the angle subtended at the arc

Now add two more line segments to create an angle at the centre (deleting the line you drew previously). Measure that angle and ask students what they notice as you change the arc.

You might want to have another look at the special case of when the arc is a semi-circle and reason why the angle at the circumference is 90°. And indeed show that this works for reflex angles too.

An alternative way of discovering this theorem is to construct this special case and then reason that because you have an **isosceles triangle** formed by **two radii**, you can show the the other angle in the isosceles triangle will be **180 – 2x the base angle**. Therefore the other angle at the centre will be 2x the base angle.

We’re not there yet, check back for Part 2 where we will go outside the circle. Oooohh….