Circle Theorems

Proof of theorem 1: The angle in a semicircle is 90°

Let us draw a line down from the angle opposite the hypotenuse to the centre. This will mean we have divided the triangle into two further triangles, each isosceles, with two sides of length r (we use r to denote radius length). This means each triangle has two angles which are the same. We can draw this below:

Circle Theorem 1 Proof, Ben Cairns, StudySmarter

Looking at the largest triangle, we know that 2x + 2y = 180° as the angles must sum to 180 °. As 2x + 2y = 180°, it follows – by dividing by two – that x + y = 90°. The angle at the circumference is given by x + y, and thus, the angle is right-angled. QED

Theorem 2: The angle at the centre is double the angle at the circumference

Here, the angle subtended by an arc at the centre is twice the angle subtended at the circumference. This is shown below. What is important to note is that it doesn't matter where the point is on the arc, as long as it is between the two unmarked angles. If this happens then the theorem will still hold.

The angle at the centre is double the angle at the circumference, Ben Cairns-StudySmarter

Proof of theorem 2: The angle at the centre is double the angle at the circumference

Let us construct the same shape, but now also construct a line from the 'x' point to the centre. This gives us two isosceles triangles with two sides of length r, and two sides of the same length. We will also have two angles in each isosceles that are the same. We will label each angle, as is shown below.

Circle Theorem 2 Proof, Ben Cairns-StudySmarter

To prove the theorem, we need to show that 2 (a + b) = c.

Using the facts that there are 180 ° in a triangle and 360 ° around a point, we can form three equations: 2a + z = 180, 2b + t = 180 and z + t + c = 360 °.

We can rearrange the first equation to z = 180-2a, and rearrange the second equation to t = 180 ° -2b.

Now we can substitute these equations into the third equation, to get 180 ° -2a + 180 ° -2b + c = 360 °.

This simplifies to c-2a-2b = 0, which can then be further simplified to 2 (a + b) = c. QED

Theorem 3: Angles from the same chord in the same segment are equal

If we have two triangles inside a circle with all three corners touching the circle, and the triangles share a side (also known as a common chord) then the third angle is the same in both triangles, as long as these third angles are in the same segment. This is shown below.

Circle Theorem 3, Ben Cairns, StudySmarter

Proof of theorem 3: Angles from the same chord in the same segment are equal

Let us first start by drawing another triangle sharing the common chord, however, this time we will connect the line to the centre. This is the same shape as seen in theorem 2, meaning we can invoke this theorem and call the angle 2x. This is shown below.

Circle Theorem 3 Proof, Ben Cairns, StudySmarter

As we have used theorem 2, it doesn't matter where we put the angle x on the arc, meaning the theorem is now proved. QED

Theorem 4: Opposite angles in a cyclic quadrilateral sum to 180°

By cyclic quadrilateral, we mean a four-sided shape, all corners of which touch a circle. When this occurs, the opposite corners in the quadrilateral will sum to 180°.

Circle Theorem 4, Ben Cairns, StudySmarter

In this case, we would have a + c = 180°, as well as b + d = 180°.

Proof of theorem 4: Opposite angles in a cyclic quadrilateral sum to 180°

Let us draw a line from each corner to the centre. As this goes from the circle to the centre, this means that this is a radius which means we have created four isosceles triangles which have pairs of matching angles. This is shown below.

Circle Theorem 4, Ben Cairns, StudySmarter

To show that opposite angles sum to 180 °, then we must show x + y + z + t = 180 °.

The angles in a quadrilateral sum to 360 °.

This means that z + y + z + t + t + x + x + y = 360 °.

This can be simplified to 2 (x + y + z + t) = 360 °, so x + y + z + t = 180 °. QED

Theorem 5: Alternate segment theorem

Suppose we drew a tangent to a circle. At the point at which the tangent touches the circle, there is a corner of a triangle. The other two corners of the triangle also lie on the circle. In this case, the angle between the tangent and the triangle is equal to the adjacent angle in the triangle. This is shown below.

Circle Theorem 5, Ben Cairns, StudySmarter

Proof of the alternate segment theorem

To prove this you only need to show this on one side, as it does not matter what triangle we choose. Construct a triangle as above, and then join each corner to the centre. Again, we have created three isosceles triangles – all with a pair of corresponding angles. We will call the angle between the tangent and triangle a. This is all shown below.

Circle Theorem 5 Proof, Ben Cairns, StudySmarter

Our aim is to show that a = z + y.

As the radius is perpendicular to the tangent at the point the tangent touches the circle (by definition), we know a + x = 90 °.

As the angles in a triangle sum to 180 °, we know 2x + 2y + 2z = 180 °, so x + y + z = 90 °.

We can rewrite our first equation as x = 90° - a, and then substitute this into the second equation, to get 90 ° -a + y + z = 90 °, which we can rearrange to a = z + y

This was our aim. QED

Theorem 6: Tangents from a point to a circle are the same length

Suppose we draw a circle and choose any point in the same plane as the circle, so long as the point is outside of the circle. Then, we can draw two tangent lines from the point to the circle. In addition to this, the distance from the point to the circle will be the same in both cases.

Circle Theorem 6, Ben Cairns-StudySmarter

So in this case, the distance AP is the same as the distance AQ.

Proof of theorem 6: Tangents from a point to a circle are the same length

Let us draw lines from the tangent point to the centre, recalling that the radius at a tangent point is perpendicular to the tangent, which then gives us:

Circle Theorem 6 Proof, Ben Cairns, StudySmarter

We can now use Pythagoras' theorem as we have right-angled triangles.

This gives us \((AO)^2 = r^2 + (AP)^2\) and \((AO)^2 = r^2 + (AQ)^2\)

Equating the two expressions for \((AO)^2\), we arrive at:

\((AP)^2 = (AQ)^2\)

\(AP = AQ\)

Length is always positive so the negative solutions are ignored.

Theorem 7: A radius which is perpendicular through a chord bisects the chord

Suppose we have any chord in a circle, and we draw a line from the radius to the circle boundary, and this line is perpendicular to the chord. In this case, the radius will bisect the chord.

Circle Theorem 7, Ben Cairns, StudySmarter

Proof of theorem 7: A radius which is perpendicular through a chord bisects the chord

Let us draw a line from O to M and also from O to N.

Circle Theorem 7 Proof, Ben Cairns, StudySmarter

As we have right-angled triangles, then we can use Pythagoras' theorem.

We get \((OM)^2 = (OA)^2 + (AM)^2\) and \((ON)^2 = (OA)^2 + (AN)^2\). As ON = r = OM, we can set both of these equal to one another, to get \((OA)^2 + (AN)^2 = (OA)^2 + (AM)^2\), from which we get \((AN)^2 = (AM)^2\) and it then follows that AN = AM.

As length is positive, then AM = AN. QED

Examples of the use of circle theorems

Example of the use of circle theorems - 1

Example of the use of circle theorems - 2