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**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 91

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: 7,613

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: 7,613

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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

Hi guys,

- look at your own responsibility

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

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

The knowledge of some things as a function of age is a delta function.

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**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 91

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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

You're welcome!

The problem is quite tricky!

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

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,613

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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

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

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

The knowledge of some things as a function of age is a delta function.

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**thedarktiger****Member**- Registered: 2014-01-10
- Posts: 91

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

Good. You can read.

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**RandomPieKevin****Member**- Registered: 2015-07-02
- Posts: 29

I don't understand how anonimnystefy got the first two ratios.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,613

hi RandomPieKevin

Welcome to the forum.

Angle BAP = ADP ( see http://www.mathisfunforum.com/viewtopic.php?id=17799 post 7 )

The ratio result then follows from "triangles ABP and DAP are similar".

Bob

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

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**dazzle1230****Member**- Registered: 2016-05-17
- Posts: 49

I still don't understand the property of the circle that makes angle BAP=ADP.

Is there another way to show that the triangles are similar?

I've tried using angle chasing, but I don't know which angles I should use to represent with the variables x and y.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,613

hi dazzle1230

This comes from one of the angle properties of a circle. Have a look at post 7 here:

http://www.mathisfunforum.com/viewtopic.php?id=17799

This relies on the other properties also proved in that thread.

I have also made a contents post for other geometry proofs which you will find here:

http://www.mathisfunforum.com/viewtopic.php?id=22506

Hope that helps,

Bob

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

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**dazzle1230****Member**- Registered: 2016-05-17
- Posts: 49

I see it now, thank you so much bob!

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 703

*Last edited by thickhead (2016-06-15 20:06:00)*

**SOMETHING IS BETTER THAN NOTHING."NOTHING" IS BETTER THAN "NONSENSE".NOTHING IS BETTER THAN NONSENSE. **

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