Answer:
Option A
Explanation:
In $\triangle $ ABC
(A) We have
$r r_{2}=r_{1}r_{3}$
$\therefore$ $\frac{\triangle}{s}.\frac{\triangle}{s-b}=\frac{\triangle}{s-a}.\frac{\triangle}{s-c}\Rightarrow\frac{(s-a)(s-c)}{s(s-b)}=1$
$\Rightarrow$ $\tan ^{2} \frac{B}{2}$=1 $\Rightarrow$ $\tan \frac{B}{2}$ = $\tan 45^{0} \Rightarrow$ B=$ 90^{0}$
$\therefore$ $b^{2} = a^{2}+c^{2}$
$\therefore$ A→ II
(B) We have
$r_{1}+r_{2}=r_{3}-r$
$\frac{\triangle}{s-a}+\frac{\triangle}{s-b}=\frac{\triangle}{s-c}-\frac{\triangle}{s}$
$\Rightarrow $ $\frac{s-b+s-a}{(s-a)(s-b)}=\frac{s-s+c}{s(s-c)}$
$\Rightarrow$ $ \frac{2s-(a+b)}{(s-a)(s-b)}=\frac{c}{s(s-c)}= \tan ^{2} \frac{C}{2}=1$
$\Rightarrow$ $\angle c=90^{0}$
$\Rightarrow$ B → III
(C) $r_{1}=r+2R$
$\frac{\triangle}{s-a}=\frac{\triangle}{s}+\frac{a}{\sin A}\Rightarrow \sin A=\frac{s(s-a)}{\triangle}$
$\Rightarrow$ $ 2\sin\frac{A}{2}\cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}=\cot \frac{A}{2}$
$\Rightarrow$ $\sin ^{2} \frac{A}{2}=\frac{1}{2}\Rightarrow \sin \frac{A}{2}=\frac{1}{\sqrt{2}}\Rightarrow \angle A=90^{0}$
$\therefore$ C → I
