 # Circle Theorems Questions, Worksheets and Revision

Level 8 Level 9

## What you need to know

There are 8 circle theorems in total, and they’re all facts about angles/lengths in particular situations all involving circles. You should be familiar with them all to the point where a) you can see when they should be used, and b) you’re able to describe which one you’ve used with appropriate language. We’ll have a look at an example below but first, here’s all 8 of them.

There is a small chance you might be asked how to prove these circle theorems, but it’s much more important that you understand how and when to use them so here we’ll be focussing on that.

Example: Below is a circle with centre C.  BD is a diameter of the circle, A is a point on the circumference. What is the size of angle CBA?

As we’re told that BD is a diameter of the circle, we know that triangle BAD is confined within the semi-circle. So, we can use the circle theorem that tells us the angle in a semi-circle is a right-angle to deduce that $\text{angle BAD } = 90\degree$.

The question is asking for angle CBA, and now we know the other two angles in the triangle we can use the fact that angles in a triangle add to 180. Subtracting from 180, we get that

$\text{Angle CBA } = 180 - 23 - 90 = 67\degree$

Example: Below is a circle with centre C. A, B, and D are points on the circumference. Angle BCD is $126\degree$ and angle CDA is $33\degree$. Find angle ABC. You must show your workings.

When a question like this tells you to show our workings, you must state what circle theorem/geometry fact you use when you use it.

There’s no way for us to immediately find the angle we want, so we’re going to try to find the other angles in the quadrilateral ABCD. The first circle theorem we’re going to use here is: the angle at the centre is twice the angle at the circumference. The angle at the centre is 126, so $\text{angle BAD } = 126 \div 2 = 63\degree$.

We now know two out of the four angles inside ABCD. To find a third, simply observe that angles around a point sum to 360, then we get that the angle at point C (the one inside ABCD) is $360 - 126 = 234\degree$. Since the angles in a quadrilateral sum to 360, if we subtract the ones we know from 360 then find the angle we’re looking for.

$\text{Angle ABC } = 360 - 33 - 63 - 234 = 30 \degree$.

### Example Questions

Firstly, recognise that since BD is a diameter, angle BAD is the angle in a semi-circle. Our circle theorems tell us that the angle in a semi-circle is a right-angle so BAD must be $90\degree$. As we now know this, we get that

$\text{Angle BAE } = 90 + 31 = 121 \degree$.

Next, we recognise that ABDE is a cyclic quadrilateral. On a related note, the second circle theorem we’re going to use is: opposite angles in a cyclic quadrilateral sum to 180. Angle BAE (which we just worked out) is opposite to angle CDE, so

$\text{Angle CDE } = 180 - 121 = 59\degree$

Then, the final step to finding angle EDA will be subtracting the size of angle CDA from that of angle CDE to get

$\text{Angle EDA } = 59 - 18 = 41\degree$.

#### Is this a topic you struggle with? Get help now.

Our first circle theorem here will be: tangents to a circle from the same point are equal, which in this case tells us that AB and BD are equal in length.

This means that ABD must be an isosceles triangle, and so the two angles at the base must be equal. In this case those two angles are angles BAD and ADB, neither of which know. Let the size of one of these angles be $x$, then using the fact that angles in a triangle add to 180, we get

$x + x + 42 = 180$.

Then, subtract 42 from both sides to get

$2x = 180 - 42 = 138$,

and divide both sides by 2 to get

$x = 69\degree$.

Now we can use our second circle theorem, this time the alternate segment theorem. This tells us that the angle between the tangent and the side of the triangle is equal to the opposite interior angle. Given that angle ADB, which is $69\degree$, is the angle between the side of the triangle and the tangent, then the alternate segment theorem immediately gives us that the opposite interior angle, angle AED (the one we’re looking for), is also $69\degree$.

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