1. Proof Circle Theorems

2. Contents • Angle at Centre • Opposite angles in cyclic quadrilateral • Alternate Segment Theorem Circle Theorems on the Web

3. O R P Prove angle at centre is twice the angle at the circumference Q Draw line QO and extend to intersect circumference at S a b Label angles <OQR is a <OQP is b S

4. O R P S Prove angle at centre is twice the angle at the circumference Q <ORQ = <OQR Base angles in Isosceles Triangle a b (OQ = OR, radii of circle) Similarly <OPQ = <OQP a b

5. O R P S Prove angle at centre is twice the angle at the circumference Q Consider <ROS a By exterior angle of triangle theorem b <ROS = 2a Similarly 2a 2b a b <POS = 2b

6. O R P S Prove angle at centre is twice the angle at the circumference Q <POR = 2a + 2b = 2(a + b) a (a+b) b <PQR = a + b <POR = 2 x <PQR 2a 2b 2(a+b) Return to Menu

7. Opposite Angles in a Cyclic Quadrilateral are supplementary Required to Prove that x + y = 180º x Draw in radii The angle at the centre is TWICE the angle at the circumference 2y 2x 2x + 2y = 360º y 2(x + y) = 360º x + y = 180º Opposite Angles in Cyclic Quadrilateral are Supplementary Return to Menu

8. Alternate Segment Theorem R Given that Q MPN is a Tangent to the circle S PS is a chord in the circle Required to Prove … that < SPN = Any angle in the major segment PQS M P N Draw diameter PR Join RS

9. Alternate Segment Theorem R (Angle in Semicircle) <PSR = 90º Q (Radius to a Tangent) <NPR = 90º <NPS + <RPS = 90º S <PRS + <RPS = 90º <PRS + <RPS = <NPS + <RPS <RPS is common M P N <PRS = <NPS Special case of Rightangle Triangle on circle’s diameter

10. Alternate Segment Theorem R Now for ANY angle in Segment PQS Q Join QS & PQ S Consider <PQS <PQS = <PRS (Angles in SAME segment) M P N Thus <SPN = Any Angle in Alternate Segment Return to Menu