Formula for CircumRadius The formula for the circumradius $r$ of a triangle $ABC$ tells me that $r={abc\over{}4\triangle}$, where the lengths of the sides are $a$, $b$, $c$.
I'm not sure, but I occasionaly got wrong values. They might have been calculation mistakes, but then I got fixated on deriving the formula for myself.
So, I took a triangle $ABC$, circumcenter $R$.
I know that:


*

*$RA=RB=RC$

*Altitudes from $R$ to sides bisect those sides.


Side $AB$ was divided into two parts each of length $c$, $BC$ into two of $a$, and $AC$ into two of $b$. Let the altitudes to $BC$ be of length $h_1$, to $AC$ be of length $h_2$, and to $AB$ of length $h_3$.
So, using the Pythagorean Theorem,
$\begin{align}r^2&=a^2+h_1^2\\r^2&=b^2+h_2^2\\r^2&=c^2+h_3^2\end{align}$
Since I know $a,b,c$, I have four variables, namely $r,h_1,h_2,h_3$. But since I have only three equations, I am unable to solve. I have tried many times, yet I cannot find any other relations.  So my primary problem is to find the fourth equation. 
Please help.
 A: We can get this formula by the following way.
Let $\angle C$ be an acute angle, $BD$ be an altitude of $\Delta ABC$ and $BE$ be a diameter of the circumcircle. 
Thus, $\Delta ABE\sim\Delta DBC$ and
$$\frac{c}{h_b}=\frac{2r}{a},$$ which gives
$$r=\frac{ac}{2h_b}=\frac{ac}{2\cdot\frac{2\Delta}{b}}=\frac{abc}{4\Delta}.$$
A: 
The marked angles are equal by the Inscribed Angle Theorem.
Note that by similar triangles,
$$
\frac hb=\frac{c/2}r\tag1
$$
Thus, the area of the triangle is
$$
A=\frac{ah}2=\frac{abc}{4r}\tag2
$$
Therefore, the circumradius is
$$
r=\frac{abc}{4A}\tag3
$$
A: You can use the fact that the area of the triangle $ABC$ will be the sum of areas of three isosceles triangles with bases $a$, $b$, $c$ and heights $h_1$, $h_2$, $h_3$. So $A_{ABC}=\frac{1}{2}(ah_1+bh_2+ch_3)$. You can use Heron's formula to find the area of the triangle $ABC$. 
Also, if you want to use the law of sines, $\frac{AB}{\sin C}=\frac{BC}{\sin A}=\frac{AC}{\sin B}=2r$ and you can express those sines using $h_1$, $h_2$, $h_3$ along with the sides and the radius to get additional equations.
A: There is a special case for Pythagorean triple where
$\space A^2+B^2=C^2.\quad$
$\text{circumradius}(c)=\dfrac{product}{4*area}
=\dfrac{ABC}{4\bigg(\dfrac{AB}{2}\bigg)}=\dfrac{ABC}{2AB}
=\dfrac{C}{2}$
