# Angles in the same segment of a circle are equal

Categories: gcse geometry

Angles in the same segment of a circle are equal. In this article we will explain what this means, and prove that it is true.

## Angles in the same segment

In this diagram, the *chord* **RQ** divides the circle into two *segments* (see parts of a circle). The larger segment (the *major* segment) is coloured magenta and the smaller one (the *minor* segment) is yellow.

The angle *a* at point **P** is called the angle in a segment (in this case it is the angle in the major segment).

Here is point **S**, at a different position on the circle:

The chord **RQ** hasn't changed, and **S** is still in the major segment, so **S** is in the *same segment* as **P**. So by the angles in a segment theorem, the angle at **S** is equal to the angle at **P**, which is *a*.

## Angles not in the same segment

Here is another case:

In this case, the chord **RQ** remains unchanged, but the point **T** is now in the minor segment (yellow) so it is not in the same segment as **P**. We will call this angle *b*, and it will not generally be equal to *a*.

In fact, angles *a* and *b* are opposite angles in a cyclic quadrilateral, so they add up to 180°.

## Proof

The proof for this theorem is similar to the proof that the angle at the centre of a circle is twice the angle at the circumference.

We construct three triangles like this, where point **O** is the centre of the circle:

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. We know that the two angles at the base of an isosceles triangle are equal.

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:

We know that the angle at **P** (angle *a*) is equal to *x + y*. So we can replace *2x + 2y* with *2a* in the earlier equation, to get:

Which means:

Dividing both sides by two gives:

The angle *z* is determined by the triangle **QRO**. If we move the point **P**, it has no effect on points **Q**, **R**, or **O**. So moving point **P** does not change angle *z*.

Since angle *a* only depends on *z* it follows that moving point **P** will not change *a*, which proves the theorem.

## 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 at the centre of a circle is twice the angle at the circumference
- Angle in a semicircle is 90 degrees
- Opposite angles in a cyclic quadrilateral add up to 180°
- Two tangents from a point have equal length

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