Properties and Solution of Triangles
10.0 Radius of the Ex-circles
10.0 Radius of the Ex-circles
An ex-circle or inscribed circle of a triangle is defined as the circle lying outside the triangle, tangent to one of its sides and
tangent to the extensions of the other two. Every triangle has three distinct excircles, each tangent to one of the triangle's sides. The center of this circle is called excentre relative to that vertex. As shown in the figure, the ex-centre of $A$ is represented as ${J_A}$, ex-centre of $B$ is represented as ${J_B}$ and ex-centre of $C$ is represented as ${J_C}$. The radius of such circles is called ex-radius.
If ${r_1}$,${r_2}$ and ${r_3}$ are the radii of the ex-circles of triangle $ABC$ opposite to the vertex $A$, $B$ and $C$. Radii of Ex-circle are given as $$\begin{equation} \begin{aligned} (i){r_1} = \frac{\Delta }{{s - a}},{r_2} = \frac{\Delta }{{s - b}},{r_3} = \frac{\Delta }{{s - c}} \\ (ii){r_1} = s\tan \frac{A}{2},{r_2} = s\tan \frac{B}{2},{r_3} = s\tan \frac{C}{2} \\ (iii){r_1} = \frac{{a\cos \frac{B}{2}\cos \frac{C}{2}}}{{\cos \frac{A}{2}}} \\ (iv){r_1} = 4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} \\\end{aligned} \end{equation} $$
These formulae can be proved in the same manner as in case of radius of incircle.
Question 20. In a triangle $ABC$, prove that $${r_1} + {r_2} + {r_3} - r = 4R = 2a{\text{cosec}}A$$
Solution: Consider LHS, we have $$\begin{equation} \begin{aligned} {r_1} + {r_2} + {r_3} - r \\ \frac{\Delta }{{s - a}} + \frac{\Delta }{{s - b}} + \frac{\Delta }{{s - c}} - \frac{\Delta }{s} \\ \Delta \left( {\frac{1}{{s - a}} + \frac{1}{{s - b}}} \right) + \Delta \left( {\frac{1}{{s - c}} - \frac{1}{s}} \right) \\ \Delta \left[ {\left( {\frac{{s - b + s - a}}{{\left( {s - a} \right)\left( {s - b} \right)}}} \right) + \left( {\frac{{s - s + c}}{{s(s - c)}}} \right)} \right] \\\end{aligned} \end{equation} $$
Using $$2s = a + b + c$$ we get $$\begin{equation} \begin{aligned} \Delta \left[ {\left( {\frac{{2s - b - a}}{{\left( {s - a} \right)\left( {s - b} \right)}}} \right) + \left( {\frac{c}{{s(s - c)}}} \right)} \right] \\ \Delta \left[ {\left( {\frac{c}{{\left( {s - a} \right)\left( {s - b} \right)}}} \right) + \left( {\frac{c}{{s(s - c)}}} \right)} \right] \\ \Delta c\left[ {\left( {\frac{{s(s - c) + (s - a)(s - b)}}{{\left( {s - a} \right)\left( {s - b} \right)s(s - c)}}} \right)} \right] \\ \Delta c\left( {\frac{{2{s^2} - s(a + b + c) + ab}}{{{\Delta ^2}}}} \right) \\ \frac{{abc}}{\Delta } \\\end{aligned} \end{equation} $$
Using $$R = \frac{{abc}}{{4\Delta }}$$we get$$4R = 2a\cos ecA$$ as $$2R = \frac{a}{{\sin A}}$$
Question 21. In a triangle $ABC$, prove that $${r_1}{r_2} + {r_2}{r_3} + {r_3}{r_1} = {s^2}$$
Solution: Consider LHS, we have $$\begin{equation} \begin{aligned} {r_1}{r_2} + {r_2}{r_3} + {r_3}{r_1} \\ \left( {\frac{\Delta }{{s - a}} \times \frac{\Delta }{{s - b}}} \right) + \left( {\frac{\Delta }{{s - b}} \times \frac{\Delta }{{s - c}}} \right) + \left( {\frac{\Delta }{{s - c}} \times \frac{\Delta }{{s - a}}} \right) \\ {\Delta ^2}\left( {\frac{{(s - c) + (s - a) + (s - b)}}{{(s - a)(s - b)(s - c)}}} \right) \\ {\Delta ^2}\left( {\frac{{3s - (a + b + c)}}{{(s - a)(s - b)(s - c)}}} \right) \\ {\Delta ^2}\left( {\frac{s}{{(s - a)(s - b)(s - c)}}} \right) \\\end{aligned} \end{equation} $$
We know that $$\Delta = \sqrt {s(s - a)(s - b)(s - c)} $$
so $$ = {s^2}$$
Question 22. In a triangle $ABC$, prove that $$r{r_1} + r{r_2} + r{r_3} = ab + bc + ca - {s^2}$$
Solution: Consider LHS $$\begin{equation} \begin{aligned} r{r_1} + r{r_2} + r{r_3} \\ = \left( {\frac{\Delta }{s} \times \frac{\Delta }{{s - a}}} \right) + \left( {\frac{\Delta }{s} \times \frac{\Delta }{{s - b}}} \right) + \left( {\frac{\Delta }{s} \times \frac{\Delta }{{s - c}}} \right) \\ = \frac{{{\Delta ^2}}}{s}\left( {\frac{1}{{(s - a)}} + \frac{1}{{(s - b)}} + \frac{1}{{(s - c)}}} \right) \\ = \frac{{{\Delta ^2}}}{s}\left( {\frac{{(s - b)(s - c) + (s - c)(s - a) + (s - a)(s - b)}}{{(s - a)(s - b)(s - c)}}} \right) \\\end{aligned} \end{equation} ...(i) $$
we know that $$\Delta = \sqrt {s(s - a)(s - b)(s - c)} $$
so $(i)$ becomes $$\begin{equation} \begin{aligned} {s^2} - bs - cs + bc + {s^2} - cs - as + ac + {s^2} - as - bs + ab \\ ab + bc + ca - s(b + c + c + a + a + b) + 3{s^2} \\ ab + bc + ca - 2s(a + b + c) + 3{s^2} \\ ab + bc + ca - 4{s^2} + 3{s^2} \\ ab + bc + ca - {s^2} \\\end{aligned} \end{equation} $$
