Yes, but it's useful to have it as a separate case.

B

Thanks

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

hi Agnishom

This is one of the five "angle properties of a circle"  **

see diagram

The angle at the centre is twice the angle at the circumference so

reflex BED = 2 x BAD

and

acute BED = 2 x BCD

As reflex BED + acute BED = 360  => BAD + BCD = 180.

This is true for both pairs of opposite angles in all cyclic quadrilaterals.

**Do you want to know the other four?

Bob

First fin BAD from the triangle ABD,and get BCD from the fact that the quadrilayeral ABCD is inscribed in a circle,which means that BAD+BCD=180°.