Segment Theorems

Circles & Segments

Circle Theorem: Angles at the circumference subtended by the same arc are equal

This theorem is also sometimes known as the same segment theorem
It states that any two angles at the circumference of a circle that are formed from the same two points on the circumference are equal
- Subtended means the equal angles are created by drawing chords from the ends of an arc
- These chords may or may not pass through the centre
This is one of the more tricky circle theorems to identify
- STEP 1
  Choose an angle on the circumference and put your index fingers on it
- STEP 2
  Use your fingers to follow the two lines that form the angle to the point where they each meet the circumference
- STEP 3
  See if there are any other lines from these two points that meet at another angle
- The two angles are equal

If giving the same segment theorem as a reason in an exam, use the key vocabulary
- ” Angles in the same segment are equal”

Exam Tip

The same segment theorem is a common circle theorem used in GCSE exam questions
Don’t be afraid of it, look for as many equal angles you can find using it and fill them in as they will help you find other angles
If you use this theorem to help you find other angles, you should still mention the same segment theorem in your reasonsa

Worked example

Find the value of θ in the diagram below, giving reasons for your answers.

There is a diameter here, splitting the circle into two semicircles.
Identify the two triangles in each semi circle and mark in the right angles using the angle in a semicircle theorem.
Find the other angles in the triangles using the rule angles in a triangle add up to 180°

In the diagram, notice how the angle θ is subtended from the same chord as the angle that is 17°.

The angle at B is 90° because it is the angle (at the circumference) in a semi circle
The angle at C is 17° because the angles in a triangle add up to 180°
θ = 17°
because angles in the same segment are equal

Alternate Segment Theorem

Circle theorem: Alternate Segment Theorem

Although one of the least obvious circle theorems to identify, this is very helpful in finding angles quickly in many questions
The Alternate Segment Theorem states that the angle between a chord and a tangent is equal to the angle in the alternate segment
You can spot this circle theorem by looking for a “ cyclic triangle”
- ie. all three vertices of a triangle lie on the circumference
- but one vertex meets a tangent – look for where 2 chords meet a tangent
To identify which angles are equal,
- Find the point where the ‘cyclic triangle’ meets the tangent and mark the angle between them
- Look for the vertex in the triangle that is opposite the marked angle at the point where the triangle meets the circumference
- Mark this angle as equal to the first angle you marked
If using this theorem as a reason for an angle in an exam, simply state the key phrase
- ” alternate segment theorem“

Exam Tip

Spotting equal angles using the alternate segment theorem can save a lot of time in the exam
- Identify if there are any triangles with all three vertices on the circumference early on
- Look to see if any of the vertices meet a tangent

Worked example

Find the value of x, stating any angle facts and circle theorems you use.

Identify the triangle in the circle with all three vertices at the circumference.

One vertex of this triangle meets a tangent at the bottom, so look for the vertex inside the triangle opposite this point and mark that angle with 2 x + 5.

Give reasons for your working as you go.

The top left angle is 2 x + 5 because of the alternate segment theorem

This angle is also subtended by the same arc as the angle at the centre.

The angle at the centre = 2(2 x + 5) because of the circle theorem
‘the angle at the centre is twice the angle at the circumference’

Form an equation.

$2 left parenthesis 2 x space plus space 5 right parenthesis space equals space 5 x space minus space 2$

Expand the brackets and solve the equation.

$table row cell 4 x space plus space 10 space end cell equals cell space 5 x space minus space 2 end cell row cell 12 space end cell equals cell space x end cell end table$

$bold italic x bold space bold equals bold space bold 12$
Using the “alternate segment theorem” and that “angles at the centre are twice angles at the circumference”

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