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•  » Will someone tell me about this circle and chord relation?

## #1 2012-05-23 22:46:12

Agnishom
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### Will someone tell me about this circle and chord relation?

In the adjoining figure,
Find angle BCD

Please give me the steps to do it

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda

## #2 2012-05-23 23:25:48

anonimnystefy
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### Re: Will someone tell me about this circle and chord relation?

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°.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #3 2012-05-23 23:35:19

Agnishom
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### Re: Will someone tell me about this circle and chord relation?

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda

## #4 2012-05-24 02:02:19

bob bundy
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### Re: Will someone tell me about this circle and chord relation?

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

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #5 2012-05-24 20:54:54

Agnishom
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### Re: Will someone tell me about this circle and chord relation?

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda

## #6 2012-05-25 17:36:11

bob bundy
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### Re: Will someone tell me about this circle and chord relation?

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

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #7 2012-05-25 17:53:22

bob bundy
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### Re: Will someone tell me about this circle and chord relation?

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

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

## #8 2012-05-25 18:04:23

Agnishom
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### Re: Will someone tell me about this circle and chord relation?

Thanks

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'The whole person changes, why can't a habit?' -Alokananda

## #9 2012-05-25 21:23:13

anonimnystefy
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### Re: Will someone tell me about this circle and chord relation?

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

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #10 2012-05-25 21:50:41

bob bundy
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### Re: Will someone tell me about this circle and chord relation?

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

B

You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
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