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(B) Given that $	riangle ABC$ is an isosceles triangle with $\angle C = 90^{\circ}$.Therefore,$AC = BC$. Let $AC = BC = a$.In $	riangle ABC$,by the

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$\sqrt{2} - 1 : \sqrt{2} + 1$

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(B) Given that $	riangle ABC$ is an isosceles triangle with $\angle C = 90^{\circ}$.Therefore,$AC = BC$. Let $AC = BC = a$.In $	riangle ABC$,by the Pythagorean theorem:$c^2 = a^2 + b^2 = a^2 + a^2 = 2a^2$$c = a\sqrt{2}$We know that the inradius $r = \frac{\Delta}{s}$ and the exradius $r_3 = \frac{\Delta}{s-c}$.Thus,$\frac{r}{r_3} = \frac{s-c}{s}$.Substituting $s = \frac{a+b+c}{2} = \frac{a+a+a\sqrt{2}}{2} = \frac{2a+a\sqrt{2}}{2} = a(1 + \frac{\sqrt{2}}{2})$:$\frac{r}{r_3} = \frac{s-c}{s} = \frac{a+b-c}{a+b+c} = \frac{a+a-a\sqrt{2}}{a+a+a\sqrt{2}} = \frac{2a-a\sqrt{2}}{2a+a\sqrt{2}} = \frac{2-\sqrt{2}}{2+\sqrt{2}}$Rationalizing the denominator:$\frac{2-\sqrt{2}}{2+\sqrt{2}} 	imes \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{4 + 2 - 4\sqrt{2}}{4-2} = \frac{6-4\sqrt{2}}{2} = 3 - 2\sqrt{2} = (\sqrt{2}-1)^2$Wait,let's re-evaluate: $\frac{2-\sqrt{2}}{2+\sqrt{2}} = \frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}(\sqrt{2}+1)} = \frac{\sqrt{2}-1}{\sqrt{2}+1}$.So,$r : r_3 = (\sqrt{2}-1) : (\sqrt{2}+1)$.

If $\triangle ABC$ is a right-angled isosceles triangle and $\angle C = 90^{\circ}$,then $r : r_3 =$

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