# Thread: Fun with the right trianlge.

1. ## Fun 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.

2. Originally Posted by ThePerfectHacker
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,

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 $\rho=\frac{a \cdot b}{a+b+c}$

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:
$A=\frac{1}{2} \cdot \rho \cdot a+\frac{1}{2} \cdot \rho \cdot b+\frac{1}{2} \cdot \rho \cdot c=\frac{1}{2} \cdot \rho \cdot (a+b+c)$

Heron set $\frac{1}{2} \cdot (a+b+c)=s$ and you get: $A=\rho \cdot s$

With a right triangle you've additional conditions:
$A=\frac{1}{2} \cdot c \cdot h=\frac{1}{2} \cdot a \cdot b=\frac{1}{2} \cdot \rho \cdot (a+b+c)$

and therefore $\rho=\frac{a\cdot b}{a+b+c}$

With $a^2=p \cdot c$ and $b^2=q \cdot c$ you can change $p \cdot q=\frac{a^2}{c} \cdot \frac{b^2}{c}=\frac{a^2 \cdot b^2}{c^2}$

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