What do you mean that wheel is round? Prove it.

Don't you just love it when someone asks you to prove something and you think why?

Don't you feel a little smug though when you can prove things.

In maths we are often asked to prove things, circle theorems being one of them.

One thing to bear in mind is that you need to be able to rearrange equations in some instances.

The proof to learn first is Angles at the centre are twice that on the circumference as it features in some of the other proofs.

Lets take a look

First of all draw a line from C through the centre. You can see that two isosceles triangles have been formed.

The triangle on the left has been labelled as w, x and x and the triangle on the right has been labelled as z, y and y.

We know

w = 180° - 2x and z = 180° - 2y

Angles around a point add up to 360°

Therefore w + z + b = 360°.

Now substitute in w and z

180 - 2x + 180 - 2y + b = 360°

Simplify

360° - 2x - 2y + b = 360°

Rearrange to get b on its own

-2x -2y + b = 0

b = 2x + 2y

B = 2(x+y)

We re labelled the diagram at the beginning with x and y in place of a. B is at the centre and we know that the angle at the centre is twice the angle on the circumference. The proof is all there.

Angle from the diameter (also known as angles in a semi circle)

Draw a line from the centre of the circle to the right angle. This has formed a radius. label each side of the radius x and y.

We know

x + x + y + y = 180°

Simplify

2x + 2y = 180°

Simplify further/ Cancel down

x + y = 90° (The angles at the right angle)

Angles in the same segment are equal

If you draw in a chord and two radii from the centre, you can see angles at the centre are twice that on the circumference.

Whatever angle is at the centre the angle at the circumference will be half. So angle a and b must be the same.

If you draw in two radii from the centre you can see that the angles in the centre rule creeps in.

We know

2x + 2y = 360°

Simplify/Cancel down

x + y = 180° (opposites sides of the quadrilateral)

Alternate segment proof

Draw radii in from the centre

Form two right angles triangles.

We know

a + x = 90° (Angles that meet at a tangent)

90 + x + y = 180°

Simplify

x + y = 90° remember a + x also equals 90°, therefore a = y (whatever x is a and y will be the same)

Look at angle B.

We know that angle B is half that of that of the angles in the centre 2y or to be more precise that angle at the centre is twice that on the circumference. (angles at centre theorem)

Now double b to get 2b = 2y.

Halve the equations gives us b = y

We already have found that a = y and can also now see that b = y. Hence proving the alternate segment theorem.

Tangents meet a radius at 90°:

Use your imagination for this one.

Draw a line in from the centre to another point on the tangent, as shown in the diagram above.

Suppose OB is not perpendicular to line AD.

Suppose OC is perpendicular to line AD

This would mean that angle OCB = 90° and OBC would be acute.

The greater side is opposite the greatest angle, using this scenario we can see that this is not the case.

OB = OE and OE > OC which is impossible.

Therefore it is only the line that meets the tangent at the circumference is perpendicular proving that OB is perpendicular to AD and therefore 90°.

Give the following a go.