Lesson video

Hi, and thank you for joining me.

My name is Dr.

Rowenson and I'll be guiding you through this lesson.

So let's get started.

Welcome to today's lesson from the unit of Circle Theorems. This lesson is called the perpendicular from the centre of a circle to a chord bisects the chord.

Now this is a circle theorem, which by the end of today's lesson we will be able to derive and also use.

Here are some previous keywords that will be useful during today's lesson, so you may want to pause the video if you want to remind yourself what they mean and press play when you're ready to continue.

The lesson is broken into two learning cycles.

In the first learning cycle, we're going to investigate this in a couple of different ways, and one of the ways will involve circular geoboards.

Then the second learning cycle, we're going to introduce the theorem more formally.

Let's start off with investigating angles with circular geoboards.

Here we have Jacob who marks a point and draws a circle around it using compasses.

He then draws a chord.

Then he's going to construct the perpendicular bisector for this chord.

When he does that, what do you think will happen? Where do you think the perpendicular bisector will go? Pause the video while you think about this and press play when you're ready to see what happens.

Let's take a look.

Jacob constructs a perpendicular bisector for this chord and he does it using a pair compasses.

He constructs this by first opening up the pair encompasses to more than half the length of the chord like so, drawing two arcs like this, drawing two arcs from the other end of the chord like this, and joining up the two points where the arcs intersect and continuing that line onwards like this.

And what we can see is we have the perpendicular bisector to the chord.

Let's break down what that means.

The line that Jacob has constructed is perpendicular to the chord and we can see that by the right angle markers on there.

So that's the perpendicular part of perpendicular bisector.

What about bisector? Well, the line bisects the chord, which means it splits it into two equal lengths.

Jacob says, "I've noticed something else.

It also goes through the centre of the circle." That means that part of this line is the radius of the circle, and that's what we can see highlighted.

Now I wonder if that always happens or if it's just a case for how Jacob has drawn it here.

Well, you can check this for yourself dear.

Take a sheet of paper and mark a point near the middle of the page and then use a per compasses to draw a large circle around the point.

Doesn't matter how big it is, but a larger one will be easier to work with.

Once you've done that, draw a record on the circle and then construct the perpendicular bisector for that chord.

Does the perpendicular bisector for your chord go for the centre of your circle? Pause while you do this and check for yourself and press play when you're ready to talk about it together.

With this particular diagram, if you construct a perpendicular bisector, it would look a bit like this, and yes, it does go through the centre, and if you've drawn yours accurately as well, yours should go through the centre too.

So let's think about this a little bit more.

We have a radius here which is a perpendicular bisector to the chord, and we have an angle mark x degrees.

What is the value of x? Pause while I write it down and press play when you're ready for an answer.

x is 90.

That angle is 90 degrees, it's a right angle.

The radius is perpendicular to the chord.

So let's put some lengths on now.

We have one length which is eight centimetres, and we have another length labelled d centimetres.

What is the value of d? Pause while I write it down and press play when you're ready for an answer.

d is equal to eight because the radius has bisected the chord, broken into two equal length.

Let's investigate this now with a circular geoboard.

Laura's using a circular geoboard that has 12 points equally spaced around a circle and a point in the centre.

She makes a triangle, then she makes a radius that goes through the midpoint of the chord to create two triangles.

She's going to calculate the size of the interior angles for each of these two triangles.

Can you anticipate what any of the angles might be? Pause the video while you think about this and press play when you're ready to continue.

Let's take a look at this together now and let's begin by just focusing on this triangle here.

Laura says, "I'll start by calculating the size of the angles in this triangle." She says the angle at the centre will be four twelfths of 360 degrees.

It's twelfths because there are 12 equally spaced points around the circle on this circular geoboard, and you can imagine that those points can be joined at the centre to create 12 equal size sectors, and it's four twelfths because the angle in this triangle that is at the centre of the circle spans four of those equal size sectors.

So when we do four twelfths of 360 degrees, we do 360 divided by 12, and then we multiply our answer by four and that would give 120 degrees for the angle in the centre.

And then she says the triangle is isosceles.

We know that because two of the sides are radii of the circle, so the two remaining angles will be equal.

We can work those out to be 30 degrees.

Now we've got those, let's put that radius back in.

Laura says the radius that goes through the midpoint of the chord bisects the angle at the centre.

Let's just break that down.

We can see that this radius goes through the midpoint of the chord by counting the pegs along this circular geoboard and seeing that's the same on either side of the chord.

So as it goes for the midpoint, it bisects the angle at the centre.

The angle at the centre was 120 degrees, so if it's bisected, it'll be broken to two equal parts which are both 60 degrees.

Laura then says, "I can now use the fact that angles and triangles sum to 180 degrees to calculate the remaining angle in each triangle." Like this, we get 90 degrees each time.

Therefore the radius that intersects the midpoint of the chord is perpendicular to it.

Another thing we can investigate is whether or not these two triangles are congruent.

What do you think? Pause the video while you consider whether or not these triangles are congruent and how you could show whether or not they are as well, and then press play when you're ready to continue.

Let's take a look at this together.

Laura says, "The radius that intersects the midpoint of the chord also bisects the chord." That means it breaks it into two equal lengths.

"Therefore the length of the line segments on either side are equal." So now we can see with these two triangles, we know all the angles are the same in each triangle and we know they have at least one length, which is the same in each triangle as well.

Laura says, "There is already enough information to prove now that the two triangles are congruent." For example, we could prove it by talking about the angle side and angle in that configuration.

But Laura says, "I could find more information about these triangles if I want to as well." For example, the hypotonus of each triangle is also equal because they are both the radii of the circle, and the two triangles share a common side as well." That means that two triangles have the same angles and the same side length, therefore they are congruent.

Let's check what we've learned there.

Here we have a circular geoboard that contains 12 equally space points around a circle and a point in the centre.

Could you please find the value of a which marks the angle at the centre.

Pause while you do it and press play when you're ready for an answer.

That angle is two twelfths of 360 degrees, which is 60.

So now we know that, could you please find the value of b? Pause the video while you do that and press play when you're ready for an answer.

The answer is 60, which we can get from subtracting 60 from 180 and then dividing by two.

Let's now draw a radius for the midpoint of this chord to break that triangle into two triangles.

We've got a new angle now, c, could you please find the value of c? Pause while you do it and press play for an answer.

The answer is 90.

We can get that from finding a sum of 60 and 30, the angles that we know in that triangle and subtracting them from 180 degrees to get 90 degrees.

Therefore, the radius is perpendicular to the chord and also bisects the chord.

Okay, it's all to you now for task A.

This task contains three questions and here is question one.

Here you've got two diagrams that show a radius intersect and a chord, and what you need to do is use a protractor to measure the size of the four angles around the intersection and also use a ruler to measure the distances along the chord from the intersection to the circle.

And once you've done that, you might wanna think about what you notice about any of the angles and lengths in either of the diagrams. Pause while you do that and press play when you're ready for question two.

And here is question two.

You've got a circle with three chords drawn on it.

Now those chords do not connect together, they're just three chords on the circle.

And what you need to do is construct the perpendicular bisector for each of those chords.

Where do those three perpendicular bisectors intersect each other? You might have a gut feeling about where they might do so, but construct a perpendicular bisector and check if you are right.

Pause while you do it and press play when you're ready for question three.

And here is question three.

You've got three diagrams which are circular geoboards containing 15 points equally spaced around a circle and a point in a centre.

And what you need to do is work out the values of the unknowns labelled a to i, and then once you've done that, think about what you notice about c degrees, f degrees, and i degrees.

Pause while you do that and press play when you are ready for answers.

Okay, let's go for the answers to question one now.

When you measure these angles, you should spot that on the left all the angles are 90 degrees, on the right they're not.

Two of them are 80 degrees and two of them are 100 degrees.

When it comes to checking the lengths, the lengths may differ depending on the size of the circles that you're working with.

For example, if it was printed at one size rather than another.

But regardless of the sizes of the circles, hopefully you spot that the lengths for the circle on the left are equal, but on the right they are unequal.

Then question two.

Once you construct the perpendicular bisector for each of the three chords, it should look something a bit like this and hopefully what you notice is that the all intersect at the centre of the circle.

Then question three, you have to find the size of some angles on circular geoboards.

Let's work through these together.

a is 72 which can work out in either of these ways on the screen, b would be 18 and c would be 90.

d, e and f could be worked out in the same ways and you should get 48, 42 and 90 respectively.

g, h and i should be 24, 66 and 90.

And then what do you notice about the angles labelled c, f and i degrees? Well, they're all right angles.

Great work so far.

Now let's move on to the next part of this lesson where we're be going to introduce the circle theorem more formally.

Here we have a diagram that shows a case with a radius is perpendicular to a chord and it also bisects the chord.

Now, Jacob says, "I wonder if a radius that is perpendicular to a chord always bisected chord." We can see it's true in this particular case, but is it true in all cases? Laura says, "Well, we could use dynamic geometry software to explore even more cases." And if you have access to this slide, there is a link at the bottom of it which opens up a GeoGebra file where you can do precisely that, and it looks something a little bit like this.

Now, while that may look convincing, all it really shows is that it's true for lots and lots of cases.

It doesn't show that it's true for every single possible case.

So how can we prove that this theorem is always true? In other words, how can we prove that whenever that line segment from the centre meets a chord at a right angle, the lengths are either side of that angle are equal? Well, Jacob and Laura have a couple of ideas.

Jacob says, "I think we could do it by considering congruent triangles." And Laura says, "I think we could do it by using Pythagoras' theorem." I wonder what they have in mind.

How could each of these approaches provide a proof.

Pause video while you think about this and press play to work through them together? Let's take a look at these together and let's start with Jacob's idea.

Jacob says, "If I can prove that these two triangles are congruent, triangle AOB and triangle BOC, then I can show that the lengths AB and BC are equal.

And that would mean that OB bisects the chord AC, because it would mean that point B will be exactly in the midpoint of AC.

So let's do that.

Triangle AOB and triangle BOC are both right angle triangles.

We can see that by the right angles marked and the fact the line segment from the centre is perpendicular to the chord.

Sides OA and OC are equal to each other because they are both radii of the circle.

Both triangles also share a common side, which is the side OB.

So triangle AOB and triangle BOC are congruent because we know they have a right angle.

We know that the hypotonus is the same in each and we know that one the sides are the same and each as well, RHS, therefore length AB and BC are both equal to each other, which means that OB bisects AC.

Well done Jacob.

Now if you want to, it might help to pause the video at this point and read through this proof again and make sure you understand where each step comes from, and then press play when you're ready to look at Laura's proof.

let's now take a look at Laura's proof.

She's labelled the length at the bottom of each triangle as a1 and a2.

And Laura says, "I'll use Pythagoras's theorem to prove that these two lengths are equal to each other, and that would mean that the chord is bisected by the perpendicular from the centre." Let's take a look at that together then.

She says the hypotenuse of each triangle are equal because they're both radii of the circle.

Let's label them both as c for good reason, Pythagoras' theorem.

She says the triangles also share a common side.

Let's label that one b, because it's one of the shorter sides of the triangle.

Then we can say that a1 is equal to the square root of c squared subtract b squared by applying Pythagoras' theorem to a triangle on the left.

We could also apply Pythagoras' theorem to a triangle on the right to say that a2 is equal to the square root of c squared subtract b squared as well.

So what we can see is that a1 and a2 are both equal to the same thing.

Therefore a1 equals a2 and the chord is bisected by the perpendicular from the centre.

So here's our circle theorem which we have derived and also now proven, that is a perpendicular from the centre of a circle to a chord bisects the chord.

Here's an example.

We have a radius and we have a chord.

The radius is perpendicular to the chord, so it bisects the chord.

And this example contains all the key properties from that circle theorem.

The radius is from the centre.

It is perpendicular to the chord and it bisects the chord.

Here's another example, but this time rather than a radius, we have just a line segment.

The line segment from the centre is perpendicular to the chord, so it bisects the chord, and we can see it fits all those properties.

It's from the centre, it's perpendicular and it bisects.

Here's a non example.

We have a radius here, but it is not perpendicular to the chord, so that means it does not bisect the chord.

Let's take a look at those properties.

Yes, this line is from the centre, but no, it's not perpendicular, so that means no, it does not bisect the chord.

Here's another non example.

This time we have a line segment that does bisect the chord.

We can see that by the hash markings, but it's not from the centre.

So that means it is not perpendicular to the chord.

Looking at those key properties again, we can see, yes, it does bisect the chord, but no, it's not from the centre, therefore, no, it is not perpendicular.

Let's check what we've learned then.

True or false? a is equal to b in this diagram.

Do you think that's true or do you think it's false and write down a reason why? Pause while you that and press play when you're ready for answers? The answer is false, and the reason why is the radius is not perpendicular to the chord, therefore it does not bisect the chord and a is not equal to b.

Here's another question.

In this diagram, a is equal to b.

Is that true or is it false? And write down a reason why.

Pause while you it and press play for an answer.

The answer is false.

Even though that line segment is perpendicular to the chord, it is not part of a line that passes through the centre of the circle, therefore it does not bisect the chord and a is not equal to b.

Here's another one.

True or false? a is equal to b in this diagram.

Pause the video while you're writing your answer down and write a reason why, and press play when you're ready for answers.

The answer is true and it's because the perpendicular to the chord from the centre bisects chord.

We can see the line segment comes from the centre, you can see it's perpendicular by the markings, therefore it must bisect the chord and the lengths a and b are equal.

Here's another question.

What is the value of x in this diagram? Pause and write it down and also write down a reason why as well, and press play when you're ready for an answer.

The answer is 90.

The reason why is because the chord is bisected by the line segment from the centre, and we can see that by the hash markings on the chord, which means that those two are the same length.

If it's bisected, it must be perpendicular because the perpendicular from the centre of a circle through a chord bisects the chord.

Okay, it's over to you now for task B.

This task contains four questions and here are questions one and two.

Pause video while do these and press play when you're ready for more questions.

And here are questions three and four.

Pause the video while you do these and press play when you're ready for answers? Okay, let's go through some answers.

In question one, you have to tick the chords that are bisected by the line segment that intersects it partway along.

What you're looking for are cases where the line segment from the centre is perpendicular to the chord, that is in these two cases here.

And in question two, you had to tick the marked angles that are right angles.

Well, once again what we're looking for here are cases where the line segment from the centre bisects chord because that means it would be a right angle, and that would be in these two cases here.

Then question three, you have to calculate the length of the chord AC.

Well, we could do that by first calculating the length of BC using Pythagoras' theorem because we have a right angle.

BC will be equal to eight centimetres and then the perpendicular from the centre of the circle to a chord bisects it, which means AC must be twice the length of BC.

Therefore AC is 16 centimetres.

Then question four, you had to calculate the value of x and give your answer accurate to three significant figures.

Well, we have a triangle which we know is a right angle triangle because we can see that the line segment from the centre bisects chord.

We can see that from the hash markings in the chord.

If we know it's a right angle triangle, we can think about things like Pythagoras and trigonometry.

Because we're trying to find a missing angle, we're going to use trigonometry.

But to use trigonometry to work out the size of that angle x, we're going to need to know two lengths on that right angle triangle, but only one of them is marked and that's the 3.

7, so where's the other length going to come from? Well, the hypotenuse of that triangle is the radius of the circle and so is the line segment that goes from the centre bisecting the chord to the circumference of the circle, and that line segment is 2.

3 plus 3.

7 units long.

That means that the length for the hypotenuse must be six units.

Now we've got that, we can use trigonometry to work out the size of the angle marked x, which is 38.

1 degrees.

Fantastic work today.

Now let's summarise what we've learned.

A theorem is a statement that can be demonstrated to be true by using accepted mathematical operations and arguments.

And today we have focused on one particular theorem and that is that a line segment from the centre of circle that is perpendicular to a chord bisects the chord.

And we can prove that to be true by either using congruent triangles or Pythagoras' theorem.

Now, theorems can be thought of as puzzles and what you are trying to do is show how to get the results.

That's your key job.

And sometimes in order to use a theorem, you may either draw a diagram or add information or extra lines to an existing diagram.

Well done today, have a great day.