1. A B P O α That is, no matter where you place point P, the angle α is always 90 0 Note: AB is the diameter of the circle whose centre is at O

2. A B P O Mark in the radius OP 2 isosceles triangles are thus formed x x y y So we can mark in angles x and y Now we add the angles in Triangle APB The angle sum is x + x + y + y = 2x + 2y So 2x + 2y = 180 0  x + y = 90 0 So angle APB is always 90 0

3. A B P O  That is, no matter where you place point P, the angle α = 2  always AB is a chord of the circle with centre O 

4. Mark in the radius OP 3 isosceles triangles are thus formed So we can mark in angles x, y and z x y A B P O y x z z Angle APB = y + z Angle AOB = 180 0 - 2x But, looking at triangle APB, 2x + 2y + 2z = 180 0  180 0 – 2x = 2(y + z)  Angle AOB = 2 x Angle APB

5. A B P1P1  That is, no matter where you place point P, the angle APB always the same – that is  =  AB is a chord of the circle, which splits the circle into 2 segments P2P2 

6. Mark in the centre O, and form the triangle AOB Let the angle AOB be 2  A B P1P1 O P2P2    Thus angle AP 1 B will be   angle at centre = double angle at circumference) Angle AP 2 B will be  for the same reason Thus angle AP 1 B = Angle AP 2 B

7. A B C That is, if AB = CD, then d 1 = d 2 AB and CD are chords of equal length D O d2d2 d1d1

8. P is the mid-point of AB, and Q is the mid-point of CD Mark in OA and OC (both are radii) A B C D O Q P Triangle OPA is congruent to triangle OQC AP = CQ  OP = OQ