Lesson video

Hello everyone and welcome to this exciting lesson on circle theorems. I'm Mr. Gratton, and get ready for this lesson where we will look at a range of different circle theorems and decide which one is appropriate to use, or maybe which combination of circle theorems are needed for any given situation.

Pause here to have a quick look at some important keywords that might be needed today.

First up, there are so many different circle theorems. Let's compare them and identify which one is appropriate for any given situation.

And Alex agrees.

There are so many different circle theorems. Which one do we use in a given situation, and what small differences about the shapes and line segments used on a circle? Means one circle theorem can be used whilst another cannot.

For example, we have a circle theorem about cyclic quadrilaterals.

Opposite angles in a cyclic quadrilateral sum to 180 degrees.

But what makes a cyclic quadrilateral? Well, all four vertices of a quadrilateral must lie on the circumference of the circle.

Unlike on these two diagrams where only three of the four vertices of the quadrilateral lie on the circumference of the circle.

The fourth point does not.

One is inside the circle whilst the other lies outside the circle.

This means these two non-examples are not cyclic quadrilaterals.

And therefore, opposite angles do not reliably sum to 180 degrees.

So pause here to think about or discuss which of these circles has a cyclic quadrilateral, and how do you know? All four vertices of circle 1 lie on the circumference of the circle.

And therefore, circle 1 has a cyclic quadrilateral.

Whilst point O on circle 2 is at the centre of the circle, not a point on its circumference.

If you know you have a cyclic quadrilateral, you can find angles using the theorem, opposite angles in a cyclic quadrilateral sum to 180 degrees.

These two angles sum to 180 degrees.

Therefore, X degrees equals 46 degrees.

A different theorem involves an angle at the centre of the circle and a second angle on its circumference.

Here are examples of this.

In every single example, we can see that the legs of the angle at the centre and the legs of the angle at the circumference meet to create two more points on the circumference of the circle.

Whilst the most familiar-looking shape that matches this criteria is the arrowhead looking quadrilateral, other shapes do exist, including a convex quadrilateral like the fourth one that may look like a cyclic quadrilateral, just with a fourth point at the centre and not on the circumference of the circle like a cyclic quadrilateral.

Remember that both the angle at the centre and the circumference must be subtended by the same arc.

In these four diagrams, we have arc DE for circles 1 and 2 and DQ for circles 3 and 4.

Furthermore, notice how in every example, except for the second, which is a little bit more complex due to the line segments overlapping, one angle is an interior angle of the quadrilateral, whilst the other angle is external to the quadrilateral.

Which means for this example, the two interior angles, DQE and DOE, are not directly related.

However, angle DQE and the external angle, reflex angle DOE, are related by the circle theorem of, the angle at the centre is twice the angle at any point on the circumference.

This is because this quadrilateral has one point at the centre of the circle, point O, and three points on the circumference of the circle, points, D, Q, and E.

We first find the reflex angle DOE at 226 degrees.

Angle Y is then half of this reflex angle at exactly 113 degrees.

Brilliant.

Let's check our understanding of these circle theorems. Pause here to find the size of angle ABC.

Select the correct circle theorem that justifies your answer.

Angle ABC is 95 degrees because the angle at the centre is twice the angle at any point on the circumference.

This is true because both angles are subtended by the same major arc, AC.

Next up, same again, but this time for angle OCB.

Pause now to find its size and justify your answer.

It's actually impossible to tell from the given information because this quadrilateral is not a cyclic quadrilateral.

Therefore, there isn't a straightforward relationship between these two opposite angles.

And lastly, find the size of angle PCB and justify your answer.

Pause now to do this.

The answer is 118 degrees because this quadrilateral is a cyclic quadrilateral, and therefore, opposite angles in a cyclic quadrilateral sum to 180 degrees.

Alex here thinks that because these two shapes look a lot alike, we can use the same circle theorem to work out both angles X and Y.

However, Izzy is a little bit more cautious.

Whilst they do both look a lot like this bow tie-looking shape, circle 3 has both angles X degrees and 36 degrees on the circumference of the circle, whereas circle 4 has one angle, angle Y degrees, at the centre of the circle instead.

So let's look at the arc that subtends all four of these angles.

For circle 3, both angles X degrees and 36 degrees are angles at the circumference of the circle that have been subtended by the same arc, arc AD.

Therefore, both angles are equal to each other.

And so X degrees equals 36 degrees.

Furthermore, rather than seeing this, that both angles are subtended by the same arc, an alternate way of viewing this is by drawing on a chord from A to D that creates a major segment of the circle.

Both angles ABD and ACD are in that same major segment, and both are angles on the circumference of the circle.

Therefore, angles in the same segment are both equal in size.

So onto circle 4.

Like we've covered before, we can use the theorem that the angle at the centre is twice the angle at the circumference.

And so the angle labelled Y degrees at the centre is double the 36 degree angle at the circumference.

And double 36 degrees is 72 degrees.

Furthermore, Izzy's comment is absolutely spot on.

We need to look at more than just the shape to know which theorem to use.

Looking at where each vertex of the shape lies is equally important.

All right, for this check, pause here to identify which chord subtends the two angles ABC and AOC.

These are the two angles.

And so chord AC subtends them both.

If it's hard to visualise, make sure to draw on some chords to help.

Right, for this same diagram, angle ABC equals 28 degrees.

Pause here to find the size of angle AOC and select the correct circle theorem that justifies your answer.

The angle is 56 degrees because the angle at the centre is twice the angle at the circumference.

We have a 28 degree angle at the circumference, and so we can double that to find the 56 degree angle at the centre.

And finally, pause here to find the size of angle CBA justifying your answer.

Angle CBA is also 23 degrees because angles in the same segment are equal.

This major segment is created by the cord CA.

This 23 degree angle is at the circumference and is equal to this other 23 degree angle, also at the circumference.

Great stuff, everyone.

On to the practise.

For question 1, calculate the size of each angle and then select from the list below the appropriate circle theorem to use for each.

You will have to use angles from previous parts to help you with each next part of the question.

Pause now for question 1.

And for question 2, each circle shows an isosceles triangle with the same chord AB in each question.

Calculate the size of each angle and write down the circle theorem that you used.

Again, you will have to use angles from previous parts of the question to help with each next part.

Pause now for question 2.

Brilliant.

Amazing work on identifying all of those circle theorems. On to the answers.

For question 1A, X equals 54, because opposite angles of a cyclic quadrilateral always sum to 180 degrees.

For question 1B, Y equals 108, because we first use the angle at the centre is twice the angle at the circumference.

To calculate the size of the angle reflex DCB, we double the 126 degree angle to get the 252 degree reflex angle.

Then to find Y, the obtuse angle, we use angles about the point, sum to 360 degrees.

So Y is the conjugate of 252, which is 108.

On to question 1C, Z equals 72 because the tangent at any point on a circle is perpendicular to the radius at that point.

So we have a quadrilateral with two interior angles of 90 degrees and another, the DCB angle, at 108 degrees.

Z is 72 because of this.

On to question 2A.

Angle CBA equals 90 degrees because the angle at the circumference of a semicircle is always a right angle.

And because the two base angles of an isosceles triangle are equal, angles ACB and angles CAB are both equal to each other at X degrees.

Therefore, the angles in this triangle are 90 degrees, 45 degrees, and 45 degrees.

This is a right-angled isosceles triangle.

For part B, we can use from part A, that the angle OAB is 45 degrees.

And because base angles in a different isosceles triangle are also equal to each other, angle OBA is also 45 degrees.

That leaves Y degrees at 90 degrees.

An alternate method is that angles at the centre are twice the angles at the circumference.

A 45 degree at the circumference is a 90 degree at the centre.

On to part C.

Angles ACB and angles ADB both equal 45 degrees because both of these angles are subtended by line segment AB.

And angles at the circumference in the same segment are equal to each other.

And since base angles in an isosceles triangle are equal, Z degrees equals 67.

5 degrees as well.

However, a super quick alternative solution is that Z degrees equals 67.

5 degrees because of the alternate segment theorem.

So we've seen what we can do to identify the correct single circle theorem to use in any given situation.

But what if we're dealing with something a little bit more complex that involves multiple circle theorems instead? Well, let's have a look.

When using multiple circle theorems to solve problems, it's especially important to communicate which circle theorems are being used and on what parts of a diagram.

We can do this by referencing explicitly the angles, line segments, and arcs involved.

For example, let's use multiple circle theorems to find angle OBC.

First, let's look at any known angles.

So this angle, BCD, is made from a tangent and a chord.

It is likely that you can find other angles using a circle theorem that involves either a tangent or a chord.

However, from the ones we know, no circle theorem can currently be used.

Sometimes it's also helpful to draw on line segments using points already on the diagram.

We have drawn one extra line segment, the chord AC.

We now have two related angles, angle BCD, made from the tangent CD and the chord BC, and this new angle BAC that has been subtended by the chord BC.

This allows us to use the alternate segment theorem.

Both these angles are therefore equal in size.

Now that we have one interior angle of this newly formed triangle, triangle ABC, let's see if we can find other angles in that triangle as well.

Since AB is a diameter of the circle, angle ACB is 90 degrees because the angle at the circumference of a semicircle is a right angle.

In other words, an angle subtended by a diameter is always 90 degrees.

Therefore, we can conclude, the angle OBC is 25 degrees because all of the interior angles of a triangle sum to 180 degrees.

Right, let's focus first on constructing more line segments to make this problem easier to solve.

We want to find the size of Angle ADC.

Pause here to identify which two helpful line segments can be drawn.

We can draw on two different radii, radii OA and OC.

If tangents are present on a diagram, it is sometimes helpful to draw on a radius from the centre to the point on the circumference that the tangent intersects.

Okay, now that we have these two line segments and knowing that angle ABC equals 41 degrees, pause here to find and justify the size of angle AOC.

We have angle ABC at the circumference and angle AOC at the centre.

Therefore, angle AOC is double angle ABC at 82 degrees.

And lastly, now that we know AOC equals 82 degrees, pause here to find and justify the size of angle ADC.

Angle ADC equals 98 degrees because the tangent at any point on a circle is perpendicular to the radius at that point.

This means that two of the interior angles of quadrilateral AOCD are 90 degrees.

Therefore, the four interior angles of this quadrilateral are 90 degrees, 90 degrees, 82 degrees, and 98 degrees.

Great stuff, everyone.

Now that you know we can use multiple circle theorems on the same diagram, let's put those skills to the test.

Pause here to write down a calculation and set of angles for each of these circle theorems on screen.

Your goal is to find the size of the angle BOD.

And for question number 2, you will need to draw one more line segment to help.

But don't worry if you draw more than one.

It might still help you to spot which one is useful.

Find the size of angle MPN just to find your answer with a circle theorem or angle fact at each stage of your working.

And for question 3, do something similar.

Draw on one more line segment in order to find angle BAO and justify your answer with theorems and facts at every stage.

Pause now for these two questions.

And finally, on to question number 4.

By justifying your answer fully, find angle VTW and pause here for this final question.

Well done, everyone, on tackling all of those questions.

On to the answers.

For question 1, for this cyclic quadrilateral, angle BAD plus angle BCD equals 180 degrees.

Therefore, angle BCD equals 133 degrees.

For the angle at the centre being twice the angle at any point on the circumference, we have reflex angle BOD being twice 133 degrees at 266 degrees.

Finally, for angles around a point sum to 360 degrees, angle BOD equals 94 degrees, the conjugate of the 266 degree angle.

For question number 2, line segment LM can be drawn to create triangle LMN.

LN is a diameter, so angle LMN is 90 degrees.

Angles at the circumference of a semicircle are always 90 degrees.

Angle NLM equals 28 degrees as it is the last interior angle of triangle LMN, and interior angles of a triangle always sum to 180 degrees.

Furthermore, angle NPM is also 28 degrees because angles in the same segment at the circumference are always equal to each other.

On to question 3.

Drawing the radius DO creates triangle COD.

We have an isosceles triangle made from two radii.

So the two base angles are both 55 degrees.

Therefore, the final angle COD is 70 degrees.

Angle BOA is also 70 degrees because vertically opposite angles are equal.

Angle OBA equals 90 degrees because the tangent and radius meet at a right angle.

Finally, angle BAO is 20 degrees because it is the final interior angle of that triangle.

Brilliant.

On to the last question.

Let's draw the chord VW.

And angle UTV equals angle VWT, equals 80 degrees because of the alternate segment theorem.

Angle TVW equals 69 degrees because it is opposite the 111 degree angle in that cyclic quadrilateral.

Finally, angle VTW equals 31 degrees because the interior angles of triangle VWT sum to 180 degrees.

Brilliant work, everyone.

Great use on all of those circle theorems in a lesson where we have looked at the differences in shape and vertex location for different triangles, quadrilaterals, and complex quadrilaterals, and how they influence the circle theorems that are appropriate.

We've also seen that we might need multiple circle theorems to tackle any given problem and that drawing extra line segments can be incredibly helpful.

Once again, thank you all so much for your efforts in today's lesson.

I have been Mr. Gratton, and you have been an absolute circle theorem champion.

Take care, everyone, and have an amazing rest of your day.