(1) In any circle the angle made by a chord AB at the centre is twice the angle made at the circumference.
In diagram 1, AOB = 2 x APB
proof: Extend PO to C (exact position is not important)
Let x = APC and y = CPB
Triangle APO is isosceles (AO = PO = radius) => PAO = x => AOC = 2x
Similarly, COB = 2y.
Thus AOB = 2x + 2y = 2(x+y) = 2 x APB
(2) Two angles on the same arc and made by the same chord will be equal.
As the angle AOB is fixed whilst AB is fixed, the point P may be moved to position Q with AQB = APB = half AOB
The proof of (1) breaks down if Q is moved so far round the circle that it moves the other side of A or the other side of B. So Q = P only whilst Q is on the same arc as P.
If Q moves to the other side of the circle use property (4)
(3) If AB is a diameter then APB = 90.
obvious as 180 = 2 x 90
(4) Opposite angles in a cyclic quadrilateral add up to 180.
Proof given in earlier post.
Note: Given any three non collinear points, it is always possible to draw exactly one circle that goes through them.
If a fourth point is placed somewhere not on the circle a quadrilateral is made that is definitely not cyclic.
But, if the point is moved onto the circle, then property (4) applies.
Property (5) in next post.