Incircle Formulae

The radius of the incircle of a  $$\Delta ABC$$  is generally denoted by r. The incenter is the point of concurrency of the angle bisectors of the angles of   $$\Delta ABC$$  , while the perpendicular distance of the incenter from any side is the radius r of the incircle:

The next four relations are concerned with relating r with the other parameters of the triangle:

\boxed{\begin{align} & \ r=\frac{\Delta }{s} \\ & \ r=(s-a)\tan \frac{A}{2}=(s-b)\tan \frac{B}{2}=(s-c)\tan \frac{C}{2}\ \\ & \ r=\frac{a\sin \frac{B}{2}\sin \frac{C}{2}}{\cos \frac{A}{2}}=\frac{b\sin \frac{C}{2}\sin \frac{A}{2}}{\cos \frac{B}{2}}=\frac{c\sin \frac{A}{2}\sin \frac{B}{2}}{\cos \frac{C}{2}}\ \\ & \ r=4\ R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\ \end{align}}

Proofs: The first of these relations is very easy to prove:

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta = {\text{area}}\;(\Delta BIC) + {\text{area}}\;(\Delta CIA) + {\text{area}}\,(\Delta AIB) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad\quad= \frac{1}{2}ar + \frac{1}{2}br + \frac{1}{2}cr\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{How?}}} \right) \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad\quad= sr \\ &\quad\Rightarrow\quad r = \frac{\Delta }{s} \\ \end{align}

To prove the second relation, we note that   $$AE=AF,BD=BF\,\,and\,\,CD=CE$$ . Thus,

From  $$\Delta AIE,$$  we have

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \frac{A}{2} = \frac{r}{{AE}} = \frac{r}{{s - a}} \\ &\Rightarrow\quad r = (s - a)\tan \frac{A}{2} \\\end{align}

Similarly, we’ll have \begin{align} r = (s - b)\tan \frac{B}{2} = (s - c)\tan \frac{C}{2}\end{align}

To prove the third relation, we have

\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = BD + CD \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad\;= \frac{r}{{\tan \frac{B}{2}}} + \frac{r}{{\tan \frac{C}{2}}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad\;= \frac{{r\sin \left( {\frac{{B + C}}{2}} \right)}}{{\sin \frac{B}{2}\sin \frac{C}{2}}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\quad\;= \frac{{r\cos \frac{A}{2}}}{{\sin \frac{B}{2}\sin \frac{C}{2}}}\qquad{(How?)} \\ &\Rightarrow\quad r = \frac{{a\sin \frac{B}{2}\sin \frac{C}{2}}}{{\cos \frac{A}{2}}} \\ \end{align}

The fourth relation follows from the third and the fact that $$a = 2R\sin A$$  :

\begin{align} r = \frac{{(2R\sin A)\sin \frac{B}{2}\sin \frac{C}{2}}}{{\cos \frac{A}{2}}} \\ \,\,\, = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\ \end{align}