# A Big Doubt (Own Theorem)

A few years ago I discovered this property and it supposed to be my own Geometry Theorem. I want to know if is it real. Have you seen a Theorem like this?

As a corollary we can say:

If we let $$a$$ and $$b$$ be the foots of a right triangle with hypotenuse $$c$$, we have

$$\dfrac {c^{2}}{a^{2}} = \sum_{n=0}^\infty (\dfrac{b}{c})^{2n}$$ And $$\dfrac{c^{2}}{b^{2}}=\sum_{n=0}^\infty (\dfrac{a}{c})^{2n}$$

Note by Hjalmar Orellana Soto
2 years, 8 months ago

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For the triangle in the above photo we know by euclidean theorem that $$a^{2}=pc$$, and $$h=\frac{ab}{c}$$. So it is true that - $$a_{1} = \frac{c_{1}h}{b}$$ - $$a_{1}=\frac{\frac{b^{2}}{c}\frac{ab}{c}}{b}=a(\frac{b}{c})^{2}$$ It's easy to see that $$c_{1}=c(\frac{b}{c})^{2}$$ by euclidean theorem. And the same with $$b_{1}$$ to get $$b_{1}=b(\frac{b}{c})^{2}$$. If you take it with the next triangles at that side you'll find that $$a_{n}=a(\frac{b}{c})^{2n}$$, $$b_{n}=b(\frac{b}{c})^{2n}$$ and $$c_{n}=c(\frac{b}{c})^{2n}$$ And with the triangles at the other side you'll find that $$a_{k}=a(\frac{a}{c})^{2k}$$, $$b_{k}=b(\frac{a}{c})^{2k}$$ and $$c_{k}=c(\frac{a}{c})^{2k}$$

- 2 years, 8 months ago

You would need to explain the set up with some words, in order to explicitly convey what you intend.

Staff - 2 years, 8 months ago

I've seen to problem of this sort. The closed form of $$a_{k}$$ wasn't given, but we had to sum a very similar series.

- 2 years, 1 month ago

My calculations in Maple show me that it's true for triangles :

(a=4, b = 8.01, c = 8.95)

(a=3, b = 4, c = 5)

(a=4.9875077, b=12.429690, c=13.392999)

- 2 years, 8 months ago

The corollary or the theorem?

- 2 years, 8 months ago

I think it's theorem, I have not seen it before.

- 2 years, 8 months ago