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If $\triangle \mathrm{ABC}$ is a right angled isosceles triangle and $\angle \mathrm{C}=90^{\circ}$, then $\mathrm{r}: \mathrm{r}_3=$
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Verified Answer
The correct answer is:
$\sqrt{2}-1: \sqrt{2}+1$
$\because A B C$ is isosceles triangle.
$\therefore A C=B C$ and $\angle C=90^{\circ}$
Let $A C=B C \Rightarrow a=b$
In $\triangle A B C$,
$\begin{aligned}
& c^2=a^2+b^2=a^2+a^2=2 a^2 \\
& \Rightarrow \quad c=a \sqrt{2}
\end{aligned}$
Now, $\frac{r}{r_3}=\frac{\frac{\Delta}{s}}{\frac{\Delta}{s-c}}=\frac{s-c}{s}$
$\begin{aligned}
& =\frac{\frac{a+b+c}{2}-c}{\frac{a+b+c}{2}}=\frac{a+b-c}{a+b+c} \\
& =\frac{a+a-a \sqrt{2}}{a+a+a \sqrt{2}}=\frac{2-\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}+1} \\
& \Rightarrow r: r_3=(\sqrt{2}-1):(\sqrt{2}+1) .
\end{aligned}$
$\therefore A C=B C$ and $\angle C=90^{\circ}$
Let $A C=B C \Rightarrow a=b$
In $\triangle A B C$,
$\begin{aligned}
& c^2=a^2+b^2=a^2+a^2=2 a^2 \\
& \Rightarrow \quad c=a \sqrt{2}
\end{aligned}$
Now, $\frac{r}{r_3}=\frac{\frac{\Delta}{s}}{\frac{\Delta}{s-c}}=\frac{s-c}{s}$
$\begin{aligned}
& =\frac{\frac{a+b+c}{2}-c}{\frac{a+b+c}{2}}=\frac{a+b-c}{a+b+c} \\
& =\frac{a+a-a \sqrt{2}}{a+a+a \sqrt{2}}=\frac{2-\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}+1} \\
& \Rightarrow r: r_3=(\sqrt{2}-1):(\sqrt{2}+1) .
\end{aligned}$
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