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Alternate Segment Theorem

Here we will learn about the alternate segment theorem, including their application, proof, and using them to solve more difficult problems.

There are also circle theorem worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is the alternate segment theorem?

The alternate segment theorem is the angle that lies between a tangent and a chord is equal to the angle subtended by the same chord in the alternate segment.

Alternate Segment Theorem image 1 1

What is the alternate segment theorem?

What is the alternate segment theorem?

Key parts of a circle needed for this theorem

Alternate Segment Theorem image 2 1

  • The circumference of the circle is the distance around the edge of the circle.
  • A chord is a straight line that meets the circumference in two places. The longest chord in a circle is the diameter.
  • The major segment is the larger part of a circle when it is enclosed by a chord and the major arc.
  • The minor segment is the smaller part of a circle when it is cut by a chord and the minor arc.
  • The tangent of a circle is a straight line that touches the circumference of the circle  with a single point of contact. The tangent is perpendicular ( 90 degrees to) the radius.

Subtended angles

An angle within a circle is created by two chords meeting at a point on the circumference.  The diagrams below show the angle subtended by arc AC from point B for two different circles.

Alternate Segment Theorem image 3 1

Top tip: The word subtend is used a lot within circle theorems so make sure you know what it means.

Proving the alternate segment theorem

To be able to prove the alternate segment theorem, you need to know the following circle theorems:

  • The angle at the centre is twice the angle at the circumference.
  • Tangents of a circle.
  • Chord of a circle.

Step
 
1
 
 
 
 
 
 
 
2
 
 
 
 
 
 
 
 
 
 
 
3
 
 
 
 
 
 
 
4
 
 
 
 
 
 
 
 
5
 
 
 
 
 
 
 
6
 
 
 
 
 
 
 
 
 
 
 
 
7
 
 
 
 
 
 
 
8

Diagram
 
Alternate Segment Theorem image 4 1
 
Alternate Segment Theorem image 4 2
 
 
 
 
 
Alternate Segment Theorem image 4 1-1
 
Alternate segment theorem 4 3
 
Alternate segment theorem 4 8
 
Alternate segment theorem 4 7
 
 
 
 
 
Alternate segment theorem 4 6
 
Alternate segment theorem 4 5
 
Alternate segment theorem 4 4

      Description
 
      Let ABC be a triangle within a circle at
      centre O. The tangent DE passes
      through the point A on the
      circumference of the circle only. The line
      AB is a chord where the point C lies on
      the major arc.
 
 
      Let us draw the two lines OA and OB.
      As they go from the centre to the
      circumference of the circle, both the
      lines OA and OB are radii
 
      This means that the lines OA and DE
      are perpendicular as the tangent to a
      circle meets the radius at 90 degrees.
 
      We will come back to this shortly.
 
 
      As OA and OB are radii, OAB is an
      isosceles triangle. This also means that
      the angles at A and B are equal. Here
      we have labelled them x.
 
 
 
 
      As AB is a chord, the perpendicular
      from the centre of a circle to a chord
      bisects the chord at 90 degrees. This
      creates two right angle triangles that
      are congruent (triangle OFA and
      triangle OFB are identical).
 
 
 
      As triangles OFA and OFB are
      congruent, the angles AOF and
      BOF are also the same. Here we have
      labelled them both y .
 
 
 
 
      Let us now inspect the angles made
      with the tangent.
 
      We know that OAE is a 90 degree
      angle from Step 2 . As OFA ios a right
      angle triangle, x+y=90^o.
 
      This means that angle FAD is also equal        to y as if we take away x from
       90^o , we get y .
 
 
      Here we now have the simplified
      diagram with the important angles
      found. The angle at O is equal to 2y as
      originally AOB was made up of y+y ,
      which is equal to 2y.
 
 
 
      As the angle at the circumference is
      twice the angle at the centre, we can
      state that the angle at C is equal to y.
      So we have angle DAB is equal to
      angle ACB.
 
 
 
      As AB is the chord, the angle at C is in
      the alternate segment to the angle
      DAB and so we can state that angles
      in the alternate segment are equal.

Tangent of a circle is one of 7 circle theorems you will need to know. You may find it helpful to start with our main circle theorems page and then look in detail at the rest.

How to use the alternate segment theorem

In order to use the alternate segment theorem:

  1. Locate the key parts of the circle for the theorem.
  2. Use other angle facts to determine one of the two angles.
  3. Use the alternate segment theorem to state the other missing angle.

Explain how to use the alternate segment theorem

Explain how to use the alternate segment theorem

Alternate segment theorem worksheet

Alternate segment theorem worksheet

Alternate segment theorem worksheet

Get your free alternate segment theorem worksheet of 20+ questions and answers. Includes reasoning and applied questions.

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Alternate segment theorem worksheet

Alternate segment theorem worksheet

Alternate segment theorem worksheet

Get your free alternate segment theorem worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE

Alternate segment theorem examples

Example 1: standard diagram

The triangle ABC is inscribed in a circle with centre O. The tangent DE meets the circle at the point A. Calculate the size of the angle ABC.

Alternate Segment Theorem example 1 1

  1. Locate the key parts of the circle for the theorem.

Alternate Segment Theorem example 1 step 1 1

Here we have:

  • The tangent DE
  • The chord AC (that meets the tangent)
  • The angle CAE = 56^o
  • The angle ABC = \theta

2Use other angle facts to determine one of the two angles.

We already know that CAE = 56^o so we do not need to use any other angle fact to determine this angle for this example.

3Use the alternate segment theorem to state the other missing angle.

ABC = 56^o as angles in the alternate segment are equal to the angle between the tangent and the associated chord.

Example 2: using angles in a triangle

Three chords meet the circumference at the points A, B, and C of the circle with centre O. DE is a tangent that meets the circle at point A. Calculate the size of angle CAE.

Alternate Segment Theorem example 2 1

Alternate Segment Theorem example 2 step 1 1


Here we have:

  • The tangent DE
  • The chord AC (that meets the tangent)
  • The angle BAC = 68^o
  • The angle ACB = 31^o
  • The angle CAE = \theta

We can use the alternate segment theorem to find the angle θ if we know the size of angle ABC. As the angle ABC = angle CAE, we can state that angle ABC is equal to \theta :


Alternate Segment Theorem example 2 step 2 1


As ABC is a triangle, we can use the fact that angles in a triangle total 180^o , and so we can find the value of ABC:


\begin{aligned}&ABC=180-(68+31) \\\\ &ABC=81^o \end{aligned}

As ABC = 81^o and angle ABC = angle CAE:


\theta=81^o

Example 3: using another circle theorem (angles in a semicircle)

The triangle ABC is inscribed in a circle with centre O. The tangent DE intersects the circle at the point C. Calculate the size of angle ACD.

Alternate Segment Theorem example 3 1

Alternate Segment Theorem example 3 step 1 1


Here we have:

  • The tangent DE
  • The chord AC (that meets the tangent)
  • The angle BCA = 17^o
  • The angle ACD = \theta

BC is a diameter as it goes through the centre at O. This means that the angle BAC is 90^o as angles in a semicircle are equal to 90^o .


The alternate segment theorem can be used if we know angle ABC as this is equal to angle ACD, which is \theta . Placing this information on the diagram, we now have:


Alternate Segment Theorem example 3 step 2 1


ABC is a triangle and we know two of the three angles so we can now work out the value of angle ABC:


\begin{aligned} &ABC=180-(90+17) \\\\ &ABC=73^o \end{aligned}

As ABC = 73^o and angle ABC = angle ACD:


\theta=73^o

Example 4: using another circle theorem (cyclic quadrilateral)

A, B, C, and D are points on the circumference of the circle with centre O. The line EF is a tangent at C. Calculate the size of the inscribed angle DCF.

Alternate Segment Theorem example 4 1

Alternate Segment Theorem example 4 step 1 1


Here we have:

  • The tangent EF
  • The chord CD (that meets the tangent)
  • The angle BDC = 72^o
  • The angle BAD= 72^o
  • The angle DCF = \theta

ABCD is a cyclic quadrilateral and so as opposite angles in a cyclic quadrilateral total 180^o , we can say that angle BCD = 180 - 95 = 85^o . Angle DBC is the required angle \theta in the alternate segment of the chord CD. Here is the updated diagram.


Alternate Segment Theorem example 4 step 2 1


We can calculate angle DBC because BCD is a triangle and angles in a triangle total 180^o . We therefore have


\begin{aligned} &DBC=180-(72+85) \\\\ &DBC=23^o \end{aligned}

As ABC = 23^o and angle DCF and angle DBC are equal angles:


\theta=23^o

Example 5: angles in parallel lines

ABC is a triangle that is inscribed in a circle with centre O. The tangent DE touches the circle at point B and is parallel to the chord AC. Calculate the size of angle ABD.

Alternate Segment Theorem example 5 1

Alternate Segment Theorem example 5 step 1 1


Here we have:

  • The tangent DE
  • The chord AB (that meets the tangent)
  • The angle ABC= 48^o
  • The angle ABD = \theta

As DE and AC are parallel and alternate angles in parallel lines are equal, we can say that angle BAC = \theta . The angle BCA is in the alternate segment to the angle ABD and so this angle is also equal to \theta .


This means that triangle ABC is an isosceles triangle as there are two angles that are equal. Here is the updated diagram showing this information:


Alternate Segment Theorem example 5 step 2 1


Now that we know ABC is isosceles, the size of angle \theta is equal to


\begin{aligned} &\theta=(180-48)2 \\\\ &\theta=66^o \end{aligned}

As angle ABD is equal to angle BCA, we can use the alternate segment theorem to state that angle ABD = 66^o .

Example 6: using another circle theorem (angle at the centre)

A circle has a centre at O and a tangent at the point C. Points A, B, and C lie on the circumference of the circle. OB and OC are radii of the circle. Calculate the size of the angle BCE.

Alternate Segment Theorem example 6 1

Alternate Segment Theorem example 6 step 1 1


Here we have:

  • The tangent DE
  • The chord BC (that meets the tangent)
  • The angle BOC = 30^o
  • The angle BCE = \theta

Angle BAC is in the alternate segment to the angle BCE so if we can calculate the size of this angle, we can then find the size of angle BCE.


Alternate Segment Theorem example 6 step 2 1


As the angle at the centre is twice the angle at the circumference, we can calculate the size of angle BAC.


\begin{aligned} &BAC=30 \div 2\\\\ &BAC=15^o \end{aligned}

As angle BAC is in the alternate segment to the angle BCE, we can use the alternate segment theorem to state that angle BCE = 15^o .

Common misconceptions

  • Tangent and radius

As the tangent meets the radius at 90 degrees, the assumption is that the angle in the alternate segment is the remainder of the angle taken from 90^o .

  • Parallel lines

Take for example the diagram below:

Alternate Segment Theorem image 1 2

The chord BC is assumed to be parallel to the tangent and so the angle ABC  is equal to the angle at the tangent. Here the angle BCA would be equal. 

Top tip: Use arrows to visualise which way the alternate angle appears:

Alternate Segment Theorem image 2 2

  • Opposite angles in a cyclic quadrilateral

The angle is taken from 180^o which is a confusion with opposite angles in a cyclic quadrilateral.

Alternate Segment Theorem image 3 2

Here, angle ABC is incorrectly calculated as 180 - 56 = 124^o .

The angle ABC = 56^o as it is in the alternate segment to the angle CAE.

  • Angle in a semicircle

The angle at the circumference is assumed to be 90 degrees although the associated chord does not go through the centre of the circle.

Practice alternate segment theorem questions

1. A, B, and C are points on the circumference of a circle with centre O. The tangent DE passes through the point A. Calculate the size of angle CAE.

 

Practice Alternate Segment Theorem question 1 1

8^o
GCSE Quiz False

90^o
GCSE Quiz False

82^o
GCSE Quiz True

98^o
GCSE Quiz False

ABC = CAE = 82^o (alternate segment theorem)

2. A, B, and C are points on the circumference of a circle with centre O. The tangent DE passes through the point A. Calculate the size of angle CAE

 

Practice Alternate Segment Theorem question 2 1

16^o
GCSE Quiz False

41^o
GCSE Quiz True

65^o
GCSE Quiz False

106^o
GCSE Quiz False

ABC = 180 \; – \; (74+65) = 41^o

 

CAE = ABC = 41^o (alternate segment theorem)

3. The points A, B, and C lie on the circle with centre O. The angle BAC = 4^o . DE is a tangent to the circle at point C. Calculate the size of angle BCE.

 

Practice Alternate Segment Theorem question 3 1

86^o
GCSE Quiz True

106^o
GCSE Quiz False

90^o
GCSE Quiz False

176^o
GCSE Quiz False

ABC = 90^o (angles in a semicircle)

 

BAC = 180\; – \; (90+4) = 86^o

 

BCE = BAC = 86^o (alternate segment theorem)

4. ABCD is a cyclic quadrilateral. Calculate the size of angle CDF.

 

Practice Alternate Segment Theorem question 4 1

55^o
GCSE Quiz False

35^o
GCSE Quiz False

99^o
GCSE Quiz False

26^o
GCSE Quiz True

ADC = 180 \; – \; 81 = 99^o

 

ACD = ADF = 55^o (alternate segment theorem)

 

CDE = 180 \; – \; (99+55) = 26^o (angles on a straight line)

5. A, B, and C are points on the circle with centre O. The chord AB is parallel to the tangent DE where DE passes through the point C. Calculate the size of angle BCE.

 

Practice Alternate Segment Theorem question 5 1

23^o
GCSE Quiz False

56.5^o
GCSE Quiz True

113^o
GCSE Quiz False

67^o
GCSE Quiz False

ABC is an isosceles triangle as AB is parallel to DE, so AC = BC.

 

BAC = (180 \; – \; 67) 2 = 56.5^o

 

BCE = BAC = 56.5^o (alternate segment theorem)

6. Below is a circle with centre O. The points A, B, and C lie on the perimeter of the circle with the tangent DE passing through the point A. Angle AOC = 42^o . Calculate the size of angle CAD.

 

Practice Alternate Segment Theorem question 6 1

96^o
GCSE Quiz False

48^o
GCSE Quiz False

84^o
GCSE Quiz False

21^o
GCSE Quiz True

ABC = 24 \times 2 = 84^o

 

CAD = ABC = 84^o (alternate segment theorem)

Alternate segment theorem GCSE questions

1. The circle with centre O has an inscribed triangle ABC. The tangent DE lies at the point C with angle BAC = 57^o . Calculate the size of angle BCE.

 

Alternate Segment Theorem GCSE question 1 1

 

(2 marks)

Show answer

BCE = 57^o

(1)

 

The alternate segment theorem

(1)

2. (a) Using the information provided below in the diagram, calculate the size of angle x.

 

Alternate Segment Theorem GCSE question 2a 1

 

(b) Hence or otherwise, calculate the size of angle y.

 

(4 marks)

Show answer

(a)

 

CDF = CBD = x = 43^o

(1)

 

The alternate segment theorem

(1)

 

(b)

 

ABD = BAH = y = 76^o

(1)

 

The alternate segment theorem

(1)

3. Using the information on the diagram below, calculate the size of angle \theta . State all of the assumptions made.

 

Alternate Segment Theorem GCSE question 3 1

 

(6 marks)

Show answer

CED = 180 – 85 = 95^o

(1)

 

Angles on a straight line total 180^o

(1)

 

ECD = 180 \; – \;(95 + 25) = 70^o

(1)

 

Angles in a triangle total 180^o

(1)

 

ACD = DAF = 70^o

(1)

 

The angle between the tangent and the chord and the angle in the alternate segment are equal or the alternate segment theorem.

(1)

Learning checklist

You have now learned how to:

  • Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results

The next lessons are

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