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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. ![]()
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Good. You can read.
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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 (2014-02-08 23:41:18)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob ![]()
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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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob ![]()
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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
The knowledge of some things as a function of age is a delta function.
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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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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
The knowledge of some things as a function of age is a delta function.
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hi Stefy,
Thanks for putting me out of my misery. I'll be able to sleep now. ![]()
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob ![]()
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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
The knowledge of some things as a function of age is a delta function.
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Yup. don't want a math problem to do THAT. ![]()
Good. You can read.
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I don't understand how anonimnystefy got the first two ratios.
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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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob ![]()
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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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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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob ![]()
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I see it now, thank you so much bob!
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Last edited by thickhead (2016-06-15 20:06:00)
{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}
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