Question

Prove that $(r_1-r)(r_2-r)(r_3-r)=4r^{2}R$.

Answer 1

Determine the proof of $(r_1-r)(r_2-r)(r_3-r)=4r^{2}R$.

Use formula:

$$\begin{gathered} ∆=\sqrt{s(s–a)(s–b)(s–c)}\mathbf{}\mathit{o}\mathit{r}\mathbf{} \\ {∆}^2=s(s–a)(s–b)(s–c) \end{gathered}$$

Solve the L.H.S part

Let $ABC$ be the triangle, with $r$ representing the inradius and $R$ representing the circumradius.

So, $r=\frac{\Delta }{s}$ where$s=\frac{(a+b+c)}{2}$

$$\begin{gathered} r_1=\frac{\Delta }{s-a} \\ {r}_2=\frac{\Delta }{s-b} \\ {r}_3=\frac{\Delta }{s-c} \end{gathered}$$

Here, $r_1,r_2,and{r}_3$ are the radii of excircles opposite to the vertices $A,B,Cof\Delta ABC$.

$$\begin{gathered} (r_1-r)(r_2-r)(r_3-r)=(\frac{\Delta }{s-a}-\frac{\Delta }{s})(\frac{\Delta }{s-b}-\frac{\Delta }{s})(\frac{\Delta }{s-c}-\frac{\Delta }{s}) \\ \Rightarrow ={\Delta }^3\left[\right(\frac{s-(s-a)}{s(s-a)}\left)\right(\frac{s-(s-b)}{s(s-b)}\left)\right(\frac{s-(s-c)}{s(s-c)}\left)\right] \\ \Rightarrow =\left(\frac{{\Delta }^3\times abc}{s^3(s-a)(s-b)(s-c)}\right) \\ \Rightarrow =(\frac{{\Delta }^2}{s^2}\times \frac{abc}{s(s-a)(s-b)(s-c)})\times \Delta \\ \Rightarrow =r^2\times \frac{abc}{\Delta } \\ \Rightarrow =r^2\times 4R \\ \Rightarrow =4R{r}^2 \end{gathered}$$

Hence, the L.H.S = R.H.S.