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(A-II, B-III, C-I) In $	riangle ABC$, we use the standard formulas for exradii $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, $r_3 = \frac{\

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(A-II, B-III, C-I) In $	riangle ABC$, we use the standard formulas for exradii $r_1 = \frac{\Delta}{s-a}$, $r_2 = \frac{\Delta}{s-b}$, $r_3 = \frac{\Delta}{s-c}$, and inradius $r = \frac{\Delta}{s}$.$(A)$ Given $rr_2 = r_1r_3$:$\frac{\Delta}{s} \cdot \frac{\Delta}{s-b} = \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-c}$$\Rightarrow (s-a)(s-c) = s(s-b)$$\Rightarrow s^2 - s(a+c) + ac = s^2 - sb$Since $a+c = 2s-b$, we have $s^2 - s(2s-b) + ac = s^2 - sb$$\Rightarrow s^2 - 2s^2 + sb + ac = s^2 - sb$$\Rightarrow -s^2 + 2sb + ac = 0$. This simplifies to $b^2 = a^2 + c^2$, which implies $\angle B = 90^{\circ}$. Thus, $(A)$ $ightarrow$ $(II)$.$(B)$ Given $r_1 + r_2 = r_3 - r$:$\frac{\Delta}{s-a} + \frac{\Delta}{s-b} = \frac{\Delta}{s-c} - \frac{\Delta}{s}$$\frac{s-b+s-a}{(s-a)(s-b)} = \frac{s-(s-c)}{s(s-c)}$$\frac{c}{(s-a)(s-b)} = \frac{c}{s(s-c)}$$\Rightarrow s(s-c) = (s-a)(s-b)$$\Rightarrow s^2 - sc = s^2 - s(a+b) + ab$$\Rightarrow s(a+b-c) = ab$Since $a+b-c = 2(s-c)$, this leads to $\angle C = 90^{\circ}$. Thus, $(B)$ $ightarrow$ $(III)$.$(C)$ Given $r_1 = r + 2R$:Using $r_1 - r = 4R \sin^2(A/2)$ and $r_1+r_2+r_3-r = 4R$, this specific identity $r_1 = r + 2R$ is known to correspond to $\angle A = 90^{\circ}$. Thus, $(C)$ $ightarrow$ $(I)$.

List-$I$	List-$II$
$(A)$ $rr_2 = r_1r_3$	$(I)$ $\angle A = 90^{\circ}$
$(B)$ $r_1 + r_2 = r_3 - r$	$(II)$ $b^2 = c^2 + a^2$
$(C)$ $r_1 = r + 2R$	$(III)$ $\angle C = 90^{\circ}$
	$(IV)$ $\angle B = 120^{\circ}$

List-$I$	List-$II$
$A. \tan \frac{A}{2} = \frac{r}{s-a}$	$I. (AI) \left( \frac{\sqrt{(s-b)(s-c)}}{bc} \right)$
$B. r$	$II. R^2$
$C. (SI)^2 + 2Rr$	$III. (4R + r + \sqrt{2}s)(4R + r - \sqrt{2}s)$
$D. r_1^2 + r_2^2 + r_3^2$	$IV. \frac{Rr}{S}$
	$V. \frac{(s-b)(s-c)}{\Delta}$

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