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The perpendicular bisector of a chord passes through the centre of a circle 236 0 28 0 62 0 B A O M OBM = 28 0 Prove this. AOB = 360 0 -236 0 = 124 0 AOB.

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Presentation on theme: "The perpendicular bisector of a chord passes through the centre of a circle 236 0 28 0 62 0 B A O M OBM = 28 0 Prove this. AOB = 360 0 -236 0 = 124 0 AOB."— Presentation transcript:

1 The perpendicular bisector of a chord passes through the centre of a circle 236 0 28 0 62 0 B A O M OBM = 28 0 Prove this. AOB = 360 0 -236 0 = 124 0 AOB is isosceles Why? OBM = ½ x (180-124) = 28 0 MOB = 62 0 Why? OMB = 180 – (62+28) = 90 0 ^ ^ ^ ^ ^ Great Marlow School Mathematics Department

2 6cm O C M A 4.25 cm M is the mid point of AC. Find the length of AC and OM. Use pythagoras to find AC. AC 2 = 6 2 + 6 2 AC 2 = 36 + 36 = 72 AC = 8.5 (1 dp) CM = 4.25 OM is perpendicular to AC. Explain! The perpendicular bisector of a chord passes through the centre of a circle OM 2 = OC 2 – CM 2 OM 2 = 6 2 – 4.25 2 OM 2 = 17.94 (2 dp) OM = 4.24 (2.dp) Great Marlow School Mathematics Department

3 The tangent to a circle meets a radius at right angles. We need to prove that PMA and PMB are congruent. B A P O PM is common PMB = PMA AM = BM PMB = PMA (SAS) PB = PA - M ^ ^ Tangents from the same point are equal. Great Marlow School Mathematics Department

4 A O B C 28 0 x z y w t AB is a diameter of a circle. ACB is an angle in a semicircle. OC is a radius. CBO = 28 0 OC = OB = OA Why? x = 28 0 Why? y = 124 0 Why? z = 56 0 Why ? w = 62 0 t = 62 0 Why? x + t = 28 0 + 62 0 = 90 0 ACB = 90 0 ^ ^ ^ The angle in a semicircle equals 90 0. Great Marlow School Mathematics Department

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6 A B O C D AOB = 2 x ACB Prove this. w + z = x + y + z (= 180 0 ) -z -z w = x + y x = y Reason? w = y + y w = 2 y x y z w s t u v s + t = t + u + v (=180 0 ) -t -t s = u + v u = v s = v + v s = 2v w = 2y s = 2v w + s = 2y + 2v = 2(y + v) AOB = 2 x ACB ^ ^ ^ ^ The angle at the centre is twice the angle at the circumference. Great Marlow School Mathematics Department

7 A B O C D x y z w s t u v

8 x y z O z = 2x z = 2y 2x = 2y x = y Prove x = y AB Angles at the circumference standing on the same arc are equal. Great Marlow School Mathematics Department

9 a + b = 180 0 c + d = 180 0 a b d c Opposite angles in a cyclic quadrilateral add up to 180 0. Great Marlow School Mathematics Department

10 Find the angles marked. O is the centre of the circle. 1 O 96 0 65 4 32 O O O O Oa d c b 23 0 17 0 57 0 d 88 0 n m 110 0 95 0 j h 76 0 e g f 70 0 Great Marlow School Mathematics Department

11 Question 1 A, B, C and D are points on the circumference of a circle centre O. AC is a diameter of the circle. Angle BDO = x°. Angle BCA = 2x°. Express, in terms of x, the size of i)angle BDA, ii)angle AOD, iii)angle ABD.(4 marks) [4] Total = 4 Great Marlow School Mathematics Department

12 ----------------------------------------- Question 1 Two tangents are drawn from a point T to a circle centre O. They meet the circle at points A and B. Angle AOB is equal to 128º. In this question you MUST give reasons for your answers. Work out the size of the angles i)APB ii)BAO iii)ABT [5] Total = 5 Great Marlow School Mathematics Department


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