# Angle at the centre of a circle is twice the angle at the circumference

Categories: gcse geometry

For any chord, we can draw two lines to form an angle at the centre of the circle, angle *b*.

We can also draw two lines to form an angle at the circumference, *a*.

The angle *b* formed at the centre of a circle is twice the angle *a* at the circumference.

Here is a video on this topic:

## Alternative shapes

If we move the point **P** around the circumference, the diagram looks different:

Due to the position of **P**, the line **PR** crosses the line **OQ**. It is important to recognise that this is still the same situation, even though it looks a little different.

The chord **RQ** still forms an angle *b* at the centre, and an angle *a* at the circumference, and it is still true that angle *b* is twice angle *a*.

Here is another slightly different arrangement:

This time the point **P** is in the *minor sector* formed by the lines **OR** and **OQ**.

The angle *b* is always in the *opposite sector* to the angle *a*. So in this case *b* is a *reflex angle* (an angle that is greater than 180°). Again, don't be confused by the different appearance of the diagram.

## Proof

The key to proving this result is to construct three triangles like this:

Notice that **OP**, **OQ** and **OR** are all radii of the circle. There is a circle theorem that tells us that 2 radii form an isosceles triangle.

This means that:

- The two angles at the base of triangle
**PQO**are equal, and we will call that angle*x*. - The two angles at the base of triangle
**PRO**are equal, and we will call that angle*y*. - The two angles at the base of triangle
**QRO**are equal, and we will call that angle*z*.

We can find the three angles of the triangle **PQR**, from the diagram:

- Angle
**P**is*x + y*. - Angle
**Q**is*x + z*. - Angle
**R**is*y + z*.

Since we know that the three angles in a triangle add up to 180°, this gives us:

This can be rearranged as:

Here is the triangle **PQR** redrawn to show the angle *a*:

We can see that the angle at *a* is equal to *x + y* from the previous diagram.

We can replace *2x + 2y* with *2a* in the earlier equation, to get:

Which means:

Now let's look at the triangle **QRO**:

For the earlier diagrams, we know that the angle at **O** is *b*, and the angles at **Q** and **R** are both equal to *z*. Since the angles in the triangle add up to 180° we can see that:

We can solve for *b*:

Compare this with the previous value for *2a*:

This proves that angle *b* is twice angle *a*.

## See also

- Parts of a circle
- Tangent and radius of a circle meet at 90°
- Two radii form an isosceles triangle
- Perpendicular bisector of a chord
- Angle in a semicircle is 90 degrees
- Angles in the same segment of a circle are equal
- Opposite angles in a cyclic quadrilateral add up to 180°
- Two tangents from a point have equal length

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