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.
Hello,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