## Presentation on theme: "© T Madas O O O O O O O The Circle Theorems. © T Madas 1 st Theorem."— Presentation transcript:

© T Madas O O O O O O O The Circle Theorems

© T Madas The perpendicular bisector of a chord passes through the centre of the circle O

O The perpendicular bisector of a chord passes through the centre of the circle

© T Madas O The perpendicular bisector of a chord passes through the centre of the circle

The shapes below have been produced by rotation. Find the centre of rotation Why does it work?

© T Madas The shapes below have been produced by rotation. Find the centre of rotation

© T Madas The shapes below have been produced by rotation. Find the centre of rotation

© T Madas The shapes below have been produced by rotation. Find the centre of rotation

© T Madas O Inscribed angles which correspond to the same arc are equal Inscribed Angle

© T Madas O Inscribed angles which correspond to the same arc are equal Does this inscribed angle correspond to the same arc?

© T Madas A central angle is twice as large as any inscribed angle which corresponds to the same arc Central Angle Inscribed Angle O

© T Madas Various Forms of the Theorem O O O O O

© T Madas O An inscribed angle which corresponds to a diameter ( or semicircle ) is a right angle

5 th Theorem

© T Madas O Tangent Tangent point A tangent and a radius drawn at any point on the circumference of the circle meet at right angles

© T Madas O The intersection of two tangents to a circle is equidistant from their points of contact. [Their angle of intersection and the central angle formed by the radii at the points of contact, are supplementary]

8 th Theorem

© T Madas O The angle formed by a chord and a tangent at one of its endpoints is equal to the inscribed angle corresponding to the same chord in the alternating segment

Circle Theorem Test

Circle Theorem Mini Test

Practice Question 1

© T Madas 95° n m 55° 85° 40° p 55° O

© T Madas 50° a b 130° 25° Can you solve this problem without a circle theorem? O

© T Madas 70° p O A B C AB = BC q r 55° 90° 35° 20°

© T Madas 30° a O b c Tangent point 60° 120°

© T Madas 57° t O r 123° 57° Can you think of another reason as to why both these angles are 57° ?

© T Madas 56° 62° w O x y z 124° 56° 62° 118°

© T Madas u 45° 160° 155° O 25° 20° 25° 135° v 20°