I Prove that the geometric mean is always the same

[Trysse]

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.

 

[andrewkirk]

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.

 

[Trysse]

@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.