Editing Incircle and excircles (section)

====Derivation of exradii formula====
Source:<ref name="Altshiller-Court 1925 79"/>

Let the excircle at side <math>AB</math> touch at side <math>AC</math> extended at <math>G</math>, and let this excircle's
radius be <math>r_c</math> and its center be <math>J_c</math>. Then <math>J_c G</math> is an altitude of <math>\triangle ACJ_c</math>, so <math>\triangle ACJ_c</math> has area <math>\tfrac12 br_c</math>. By a similar argument, <math>\triangle BCJ_c</math> has area <math>\tfrac12 ar_c</math> and <math>\triangle ABJ_c</math> has area <math>\tfrac12 cr_c</math>. Thus the area <math>\Delta</math> of triangle <math>\triangle ABC</math> is
:<math display=block>\Delta = \tfrac12 (a + b - c)r_c = (s - c)r_c</math>.

So, by symmetry, denoting <math>r</math> as the radius of the incircle,
:<math display=block>\Delta = sr = (s - a)r_a = (s - b)r_b = (s - c)r_c</math>.

By the [[Law of Cosines]], we have
:<math display=block>\cos A = \frac{b^2 + c^2 - a^2}{2bc}</math>

Combining this with the identity <math>\sin^2 \! A + \cos^2 \! A = 1</math>, we have
:<math display=block>\sin A = \frac{\sqrt{-a^4 - b^4 - c^4 + 2a^2 b^2 + 2b^2 c^2 + 2 a^2 c^2}}{2bc}</math>

But <math>\Delta = \tfrac12 bc \sin A</math>, and so
:<math display=block>\begin{align}
  \Delta &= \tfrac14 \sqrt{-a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2 c^2 + 2 a^2 c^2} \\[5mu]
         &= \tfrac14 \sqrt{(a + b + c)(-a + b + c)(a - b + c)(a + b - c)} \\[5mu]
         & = \sqrt{s(s - a)(s - b)(s - c)},
\end{align}</math>

which is [[Heron's formula]].

Combining this with <math>sr = \Delta</math>, we have
:<math display=block>r^2 = \frac{\Delta^2}{s^2} = \frac{(s - a)(s - b)(s - c)}{s}.</math>

Similarly, <math>(s - a)r_a = \Delta</math> gives
:<math display=block>\begin{align}
  &r_a^2 = \frac{s(s - b)(s - c)}{s - a} \\[4pt]
  &\implies r_a = \sqrt{\frac{s(s - b)(s - c)}{s - a}}.
\end{align}</math>