A B

P

O

α

That is, no matter where you place point P, the angle α is always 900

Note: AB is the diameter of the circle whose centre is at O

A B

P

O

Mark in the radius OP

2 isosceles triangles are thus formedx

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 = 1800 x + y = 900

So angle APB is always 900

A

B

P

O

That is, no matter where you place point P, the angle α = 2always

AB is a chord of the circle with centre O

Mark in the radius OP

3 isosceles triangles are thus formed

So we can mark in angles x, y and z

x

y

AB

P

O

y

xz

z

Angle APB = y + z

Angle AOB = 1800 - 2x

But, looking at triangle APB, 2x + 2y + 2z = 1800

1800 – 2x = 2(y + z)

Angle AOB = 2 x Angle APB

A

B

P1

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

P2

Mark in the centre O, and form the triangle AOB

Let the angle AOB be 2

AB

P1

O

P2

Thus angle AP1B will be angle at centre = double angle at circumference)

Angle AP2B will be for the same reason

Thus angle AP1B = Angle AP2B

A

B

C That is, if AB = CD, then d1 = d2

AB and CD are chords of equal length

D

O

d2

d1

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