9.0 Radius of Incircle

In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in a triangle. It touches all the three sides of a triangle. The center of this circle is called in center and radius called radius of in a circle. In the center is found by the intersection of all the three internal angle bisectors.

Radius of incircle is given as $$\begin{equation} \begin{aligned} (i)r = \frac{\Delta }{s} \\ (ii)r = (s - a)\tan \frac{A}{2} = (s - b)\tan \frac{B}{2} = (s - c)\tan \frac{C}{2} \\ (iii)r = \frac{{c\sin \frac{B}{2}\sin \frac{A}{2}}}{{\cos \frac{C}{2}}} \\ (iv)r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\\end{aligned} \end{equation} $$

Proof: $(i)$ Given $\Delta ABC$ with incircle having center at $O$, we note that Area ($\Delta ABC$ ) = Area ($\Delta AOB$ ) + Area ($\Delta BOC$ ) + Area ($\Delta COA$ ). $$\begin{equation} \begin{aligned} \Delta ABC = \frac{1}{2} \times (AB \times r) + \frac{1}{2} \times (BC \times r) + \frac{1}{2} \times (CA \times r) \\ \Delta ABC = \frac{1}{2} \times r \times (AB + BC + CA) \\ \Delta ABC = \frac{1}{2} \times r \times 2s \\\end{aligned} \end{equation} $$.

$$r = \frac{\Delta }{s}$$

$(ii)$ We have to prove $$r = (s - a)\tan \frac{A}{2}$$

We know that $$\tan \frac{A}{2} = \frac{\Delta }{{s(s - a)}}$$

so RHS becomes $$\begin{equation} \begin{aligned} (s - a)\frac{\Delta }{{s(s - a)}} \\ = \frac{\Delta }{s} \\\end{aligned} \end{equation} $$

and we know that $$r = \frac{\Delta }{s}$$

therefore, $$r = (s - a)\tan \frac{A}{2}$$

Similarly, we can prove others.

$(iii)$ To prove this $$r = \frac{{a\sin \frac{A}{2}\sin \frac{B}{2}}}{{\cos \frac{C}{2}}}$$

Consider RHS $$\frac{{a\sin \frac{A}{2}\sin \frac{B}{2}}}{{\cos \frac{C}{2}}}$$

And we know that $$\begin{equation} \begin{aligned} \sin \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{bc}}} \\ \sin \frac{B}{2} = \sqrt {\frac{{(s - a)(s - c)}}{{ac}}} \\ \cos \frac{C}{2} = \sqrt {\frac{{s(s - c)}}{{ab}}} \\\end{aligned} \end{equation} $$

Putting these values in RHS, we get $$\begin{equation} \begin{aligned} \frac{{c \times \sqrt {\frac{{(s - b)(s - c)}}{{bc}}} \times \sqrt {\frac{{(s - a)(s - c)}}{{ac}}} }}{{\sqrt {\frac{{s(s - c)}}{{ab}}} }} \\ \sqrt {\frac{{(s - a)(s - b)(s - c)}}{s}} \\ = \frac{\Delta }{s} \\\end{aligned} \end{equation} $$

And we know that $$r = \frac{\Delta }{s}$$

therefore, $$r = \frac{{c\sin \frac{A}{2}\sin \frac{B}{2}}}{{\cos \frac{C}{2}}}$$

$(iv)$ To prove this $$r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$$

Consider RHS,

$$r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$$

As we know that $R$ is circumradius which is given as $$R = \frac{{abc}}{{4\Delta }}$$ and using trigonometric half angles formula of sine function which is $$\begin{equation} \begin{aligned} \sin \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{bc}}} \\ \sin \frac{B}{2} = \sqrt {\frac{{(s - a)(s - c)}}{{ac}}} \\ \sin \frac{C}{2} = \sqrt {\frac{{(s - a)(s - b)}}{{ab}}} \\\end{aligned} \end{equation} $$

Putting all these values in RHS, we get $$\begin{equation} \begin{aligned} 4 \times \frac{{abc}}{{4\Delta }} \times \sqrt {\frac{{(s - b)(s - c)}}{{bc}}} \times \sqrt {\frac{{(s - a)(s - c)}}{{ac}}} \times \sqrt {\frac{{(s - a)(s - b)}}{{ab}}} \\ \frac{{(s - a)(s - b)(s - c)}}{\Delta } \\\end{aligned} \end{equation} ...(i) $$

We know that $$\Delta = \sqrt {s(s - a)(s - b)(s - c)} $$

Squaring both sides we get,

$$\begin{equation} \begin{aligned} {\Delta ^2} = s(s - a)(s - b)(s - c) \\ (s - a)(s - b)(s - c) = \frac{{{\Delta ^2}}}{s} \\\end{aligned} \end{equation} $$

Putting $(s - a)(s - b)(s - c) = \frac{{{\Delta ^2}}}{s}$ in $(i)$ we get

$$\frac{\Delta }{s}$$

And we know that $$r = \frac{\Delta }{s}$$

therefore, $$r = 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$$

Question 19. In a triangle $ABC$, prove that $$4\left( {\frac{s}{a} - 1} \right)\left( {\frac{s}{b} - 1} \right)\left( {\frac{s}{c} - 1} \right) = \frac{r}{R}$$