The Alternate Segment Theorem (Cambridge (CIE) IGCSE International Maths)

Revision Note

Written by: Naomi C

Reviewed by: Dan Finlay

Did this video help you?

Alternate Segment Theorem

Circle theorem: The Alternate Segment Theorem

The angle between a chord and a tangent is equal to the angle in the alternate segment

Alternate Segment Theorem, IGCSE & GCSE Maths revision notes

To spot this circle theorem on a diagram
- look for a cyclic triangle
  - where all three vertices of the triangle lie on the circumference
- one vertex of the triangle meets a tangent
To identify which angles are equal
- mark the angle between the tangent and the side of the cyclic triangle
- the angle inside the triangle at the corner opposite the side of the triangle that forms the first angle is the equal angle
When explaining this theorem in an exam you can just say the phrase:
- The Alternate segment theorem

Examiner Tips and Tricks

Look for cyclic triangles and tangents in busy diagrams
- Questions involving the alternate segment theorem frequently appear in exams!

Worked Example

Find the value of $x$ .

A circle with centre, O, has three points on the circumference, P, Q and R. A tangent meets the circumference of the circle at point R. PQR and QOR are triangles. Angle QOR = (5x-2) and the angle between QR and the tangent is (2x+5).

Identify the cyclic quadrilateral (triangle in the circle with all three vertices at the circumference)

One vertex of this triangle meets a tangent at point R
The angle between one of its sides (QR) and the tangent is given
Find the angle inside the triangle, opposite to the same side (QR)

Angle between QR and the tangent = Angle RQP = $(2 x + 5)$
Alternate segment theorem

Same circle diagram with the angle RPQ marked as (2x+5).

Notice that angle RPQ and angle ROQ both come from the same two points on the circumference

Angle ROQ = $2 (2 x + 5)$
Angle at the centre is twice the angle at the circumference

Form an equation using the two expressions for angle ROQ

$2 (2 x + 5) = 5 x - 2$

Expand the brackets and solve

$\begin{array}{rcl} 4 x + 10 & = & 5 x - 2 \\ 12 & = & x \end{array}$

$x = 12$

Last updated: 17 October 2024

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Previous:Angles in the Same SegmentNext:Perimeter