The other day I saw something interesting,

Take any right triangle. Then the products of the segments of the hypotenuse divided by the inscribed circle is its area.

Printable View

- Jan 23rd 2006, 11:48 AMThePerfectHackerFun with the right trianlge.
The other day I saw something interesting,

Take any right triangle. Then the products of the segments of the hypotenuse divided by the inscribed circle is its area. - Jan 24th 2006, 10:42 AMearbothQuote:

Originally Posted by**ThePerfectHacker**

I'm a little bit puzzled about your question: What are the segments of the hypotenuse and what exactly do you take as divisor? The radius, the perimeter or the area of the inscribed circle?

I've attached a drawing of a right triangle so you can check which names I use for the variables.

Heron of Alexandra had found that in a right triangle is

The three bisectors of a triangle divide it into three smaller triangles. The height in every of these smaller triangles is always ρ. That's true with every triangle.

So you get:

Heron set and you get:

With a right triangle you've additional conditions:

and therefore

With and you can change

Now you want to divide this term by a part of the inscribed circle. But I cann't help you any further because I don't know which part you want to take. But maybe you can use my remarks to get a little bit ahead.

Bye