hi Agnishom

(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.

Bob