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Topic review (newest first)

bob bundy
2012-05-25 21:50:41

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


2012-05-25 21:23:13

Isn't (3) just a special case of (1)?

2012-05-25 18:04:23


bob bundy
2012-05-25 17:53:22

Property (5)

The angle between a tangent and a chord is equal to the angle made by the chord on the opposite side of the circle.

In the diagram, TB is a tangent to the circle at B.  TBA = APB

proof;  AOB = 2x + 2y (see proof above for definition of x and y)

=> OBA = 90 - (x+y)

But OBT = 90  =>  TBA = x + y


bob bundy
2012-05-25 17:36:11

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.


2012-05-24 20:54:54

Thanks Bob;
Now I understand
Will you kindly tell me the other properties?

bob bundy
2012-05-24 02:02:19

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


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?


2012-05-23 23:35:19

How do you know that BAD+BCD=180 degrees?
What is the property being used?

2012-05-23 23:25:48

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.

2012-05-23 22:46:12

In the adjoining figure,
Find angle BCD

Please give me the steps to do it

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