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**Alternate Segment Theorem**

Teach GCSE Maths Alternate Segment Theorem

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**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

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**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.

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**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.

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**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.

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**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.

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**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

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**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 )

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**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 )

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**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 )

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**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 )

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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.

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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.

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**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 )

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**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

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**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. )

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