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**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 83

Let \overline{PA} and \overline{PC} be tangents from P to a circle. Let B and D be points on the circle such that B, D, and P are collinear. Prove that AB \cdot CD = BC \cdot DA.

This is driving me CRAZY and the moment.

several swearing smileys should make me feel better.

Good. You can read.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,532

HI thedarktiger,

I know the feeling. I still have some problems posted over a year ago on my 'to-do' list. I remain optimistic about them.

EVEN LATER EDIT:

I thought this was going to come out easily; but it hasn't. Nevertheless I think the approach is the right one ... I've just got in a muddle with my similar triangles.

Your problem is a little like a theorem about intersecting chords so I've used a similar method. I haven't used the tangents at all; whoops!I haven't changed my mind about this. If you pick any two points (not diametrically opposite) you can construct tangents that will intersect at P. As P doesn't feature in the required equation, I cannot see why you cannot treat P as any point just to determine B and C. So all the 'action' should be able to take place inside the circle. As I type this, I'm already thinking I'm wrong so maybe I'll have a new approach soon )

I have now explored what happens if P isn't as specified and the property fails to hold. Conclusion: what I wrote earlier is wrong so I shall delete it and go back to the drawing board.

Bob

*Last edited by bob bundy (2014-02-08 23:41:18)*

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,532

hi,

It has driven me crazy too. I have drawn and re-drawn the diagram with varied emphasis and called angles x, y and z and then found every other angle in terms of these. I have experimented with similar triangles until my head feels like bursting. And still no joy.

With my geometry program I have verified that it is, indeed, true. But I am at a loss as to how to prove it.

Very sorry. Maybe tomorrow ..........

Bob

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

Hi guys,

- look at your own responsibility

*Last edited by anonimnystefy (2014-02-09 17:56:04)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 83

Thanks for the anwser, anonimnystefy! I think my head has already burst. That problem was plain evil.

I'm sure anybody named bob will agree. aaaaaaggggghhhhhh.

Good. You can read.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

You're welcome!

The problem is quite tricky!

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,532

hi Stefy,

Thanks for putting me out of my misery. I'll be able to sleep now.

Bob

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

You're welcome, but I sure am hoping you are using that as a figure of speech only.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 83

Yup. don't want a math problem to do THAT.

Good. You can read.

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