# Alternate Segment Theorem

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Alternate Segment Theorem
Teach GCSE Maths Alternate Segment Theorem

Alternate Segment Theorem
Teach GCSE Maths Alternate Segment Theorem "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" © Christine Crisp

major segment minor segment
When we draw a chord joining 2 points on the circumference of a circle, we form 2 segments. O major segment minor segment The major segment is the larger one.

P is formed by lines from A and B.
If we now draw a tangent at one end of the chord . . . we make an acute angle between the tangent and the chord, TAB O P x If we also draw an angle in the major segment . . . B the alternate segment theorem says x A TAB = APB T P is formed by lines from A and B. P can be anywhere on the circumference of the major segment.

P is formed by lines from A and B.
If we now draw a tangent at one end of the chord . . . we make an acute angle between the tangent and the chord, TAB O P x P x If we also draw an angle in the major segment . . . B the alternate segment theorem says x A TAB = APB T P is formed by lines from A and B. P can be anywhere on the circumference of the major segment.

This property of angles is called the alternate segment theorem.
The same is true for the obtuse angle but it is equal to the angle in the minor segment. O This property of angles is called the alternate segment theorem. The word “alternate” is used because the equal angles are on different sides of the chord. y y We’ll use an example to see why the alternate segment theorem is true.

TAC = 90 D C CAB = 40 90 - 50 B ABC = 90 ACB = 50 A T ADB = 50
Each time the animation pauses, decide with your partner what the reason is for the given angle. We want to show that angle ADB = 50 TAC = 90 O e.g. D C Angle between tangent and radius 50 50 CAB = 40 90 - 50 B ABC = 90 Angle in a semi-circle 40 ACB = 50 50 A 3rd angle of triangle T ADB = 50 Angles in the same segment

TAB = x TAC = 90 CAB = a = 90 - x CBA = 90 ACB = 180 – 90 – a
To prove the alternate segment theorem we use the same steps as in the example. O A B T D C TAB = x Let TAC = 90 ( angle between tangent and radius ) CAB = a = 90 - x a CBA = 90 x ( angle in a semi-circle ) ACB = 180 – 90 – a ( 3rd angle of triangle )

TAB = x TAC = 90 CAB = a = 90 - x CBA = 90 ACB = 180 – 90 – a
To prove the alternate segment theorem we use the same steps as in the example. O A B T D C TAB = x Let TAC = 90 ( angle between tangent and radius ) CAB = a = 90 - x a CBA = 90 x ( angle in a semi-circle ) ACB = 180 – 90 – a ( 3rd angle of triangle ) = 90 – ( 90 - x )

TAB = x TAC = 90 CAB = a = 90 - x CBA = 90 ACB = 180 – 90 – a
To prove the alternate segment theorem we use the same steps as in the example. O A B T D C TAB = x Let x x TAC = 90 ( angle between tangent and radius ) CAB = a = 90 - x a CBA = 90 x ( angle in a semi-circle ) ACB = 180 – 90 – a ( 3rd angle of triangle ) = 90 – ( 90 - x ) = 90 – 90 + x = x ADB = x ( angle in the same segment )

TAB = x TAC = 90 CAB = a = 90 - x CBA = 90 ACB = 180 – 90 – a
To prove the alternate segment theorem we use the same steps as in the example. O A B T D TAB = x Let x TAC = 90 ( angle between tangent and radius ) CAB = a = 90 - x CBA = 90 x ( angle in a semi-circle ) ACB = 180 – 90 – a ( 3rd angle of triangle ) = 90 – ( 90 - x ) = 90 – 90 + x = x ADB = x ( angle in the same segment )

SUMMARY The theorems involving tangents are: The angle between a tangent and the radius at the point of contact is always 90. The tangents from an external point are equal in length. The angle between a tangent and chord equals the angle in the alternate segment.

SUMMARY When solving problems we might also use the following: The perpendicular from the centre to a chord bisects the chord. The angle at the centre is twice the angle at the circumference. Angles in the same segment are equal. The angle in a semi-circle is always 90. The sum of the opposite angles of any cyclic quadrilateral is 180.

e. g. In the following, the red line is a tangent
e.g. In the following, the red line is a tangent. Find a and b giving the reasons. a b Solution: a = 115 115 (alternate segment ) b = 2a = 230 ( angle at the centre = twice the angle at the circumference )

x = 65 y = 130 z = 25 O A C B T 65 65 EXERCISE
1. In the following, O is the centre of the circle and TB is a tangent. Find x, y and z giving the reasons. O A C B T Solution: x 65 x = 65 ( angle in the alternate segment ) 25 z y = 130 130 y ( angle at the centre = twice the angle at the circumference ) z = 25 ( isosceles triangle: OC and OB are radii ) 65

B T C TA = TB A TBA = 65 BCA = 65 ADB = 115 65 50 115 65
EXERCISE 2. In the following, find angles TBA, BCA and ADB giving the reasons. 65 B T 50 D 115 O Solution: 65 C TA = TB ( tangents from one point ) A TBA = 65 ( triangle TAB is isosceles ) BCA = 65 ( angle in the alternate segment ) ADB = 115 ( opposite angles of cyclic quad. )