Prove that the geometric mean is always the same

  • #1
Trysse
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1669807413784.png

Given are a fixed point ##P## and a fixed circle ##c## with the radius ##r##. Point ##P## can be anywhere inside or outside the circle. I now draw two arbitrary lines ##l_1## and ##l_2## through the point ##P## in such a way, that both lines intersect with the circle ##c## in two distinct points. I name the points ##C_1## and ##C_2## for the line ##l_1##, respectively ##C_3## and ##C_4## for the second line ##l_2##. This gives me two sets of distances. For ##l_1## these are the distances ##PC_1## and ##PC_2##. For ##l_2## these are the distances ##PC_3## and ##PC_4##

For reference see the GeoGebra model: https://www.geogebra.org/classic/xvgereug

According to the GeoGebra model, the product of the two distances always has the same value for a given point ##P## and a given circle ##c## regardless of the orientation of the two lines.

$$PC_1 \cdot PC_2 = PC_3 \cdot PC_4$$

If I draw the line ##l_1## in such a way, that the distances ##PC_1=PC_2## this length is the square root of any two distances ##PC_3## and ##PC_4##. I.e.,

$$PC_1=PC_2 \Rightarrow PC_1=\sqrt {PC_3 \cdot PC_4}$$

So I have two questions:
1)
How can I prove, that the geometric mean (or the product) of the two distances ##PC_1## and ##PC_2## is always the same for any line through the point ##P## that intersects the circle ##c## in two points?

2)
I was thinking, that the value of the geometric mean of these two distances is somehow the geometric mean of all distances the point ##P## has to all points that lie on the circle ##c##. However, I found this post on StackExchange that argues otherwise. My intuitive approach was to calculate the following:

$$\sqrt[4] {PC_1 \cdot PC_2 \cdot PC_3 \cdot PC_4}$$

As I add ever more lines ##l_n## so I get more distances ##PC_{2n-1}## and ##PC_{2n}## that I can add in pairs to calculate the geometric mean, the result will always be the same. What is wrong with my approach? For the StackExcnage post, see here:

https://math.stackexchange.com/ques...stances-from-point-to-every-point-on-a-circle

In the Geogebra model, you can set the point ##P## with drag-and-drop and then change the lines' orientation by moving the pink points.

This is not homework.
 
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Answers and Replies

  • #2
andrewkirk
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What you are trying to prove is the "Intersecting Chords Theorem". There's a nice, short, simple proof here.
The result follows from demonstrating that triangles ##PC_1C_4## and ##PC_3C_2## are similar, because angles on the circle at the circumference standing on the same chord are equal.
 
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  • #3
Trysse
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@andrewkirk thanks. That is what I was looking for.. I would have never found this theorem on Google. I was always searching for "geometric mean, & circle & proof" but never for "chords".... Thanks for pointing me in the right direction.
 

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