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Teach GCSE Maths Alternate Segment Theorem

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Teach GCSE Maths Alternate Segment Theorem © Christine Crisp "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"

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

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

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O The same is true for the obtuse angle but it is equal to the angle in the minor segment. y y 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. Well use an example to see why the alternate segment theorem is true.

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

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

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

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

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

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

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

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O A C BT 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. x y z Solution: x = 65 ( angle in the alternate segment ) y = 130 ( angle at the centre = twice the angle at the circumference ) z = 25 ( isosceles triangle: OC and OB are radii )

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

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