1. 2. Chords are of equal length if and only if they are equidistant from the centre of circle.

2. Proof OE=OF (given) AO=BO (radii of the circle are the same) OC=OD (radii of the circle are the same) Therefore, triangle AOC is congruent to triangle BOD. Hence, we can say that AC=BD. A B C D O E F

3. 3. Angle at the centre of a circle is twice any angle at the circumference subtended by the same arc.

4. Proof OA=OC=OD (radii of a circle are the same) ADO=DAO (base angles of an isosceles triangle) AOE=ADO+DAO (ext. angle of a triangle=sum of int. angles of the triangle) AOE=2DAO=2ADO B A C D O

5. Proof Similarly, ODC=OCD (base angles of an isosceles triangle) BOC=2ODC=2OCD(ext. angle of a triangle=sum of int. angles of the triangle) Therefore, AOC=2ADO+20DC =2ADC B A CD O

6. 5. Angels in the same segment (subtended by the same arc) of a circle are equal.

7. Proof AOC = 2 ADC AOC = 2 AEC (Property 3) ADC = AEC B A C D O E

8. 6. In a cyclic quadrilateral, the opposite angels are supplementary ie. Their sum is 180 ˚

9. Proof AOC = 2 ADC AOD+COD = 2 ABC (Property 3) 2ADC+2ABC = AOC+AOD+COD = 360˚ ADC+ABC = 180 ˚ B A C D O

10. 7. The exterior angel of a circle quadrilateral is equal to the interior opposite angel

11. Proof ABC+ADC = 180˚ (Property 6) ABC+CBE = 180˚ CBE = ADC B A C D O E

12. 8. The tangent is perpendicular to the radius drawn to the point of contact.

13. Proof If OB was not perpendicular to AB, draw a perpendicular line OC, cut the circle at D. OC = OD + DC OD = OB(radii of the circle are the same) OC is longer than OB O A BC D

14. Proof The perpendicular line drawn from a point is the shortest one OC could be longer than OB if OB was not perpendicular to AB as what we assumed OB is perpendicular to AB O A BC D