In triangle ABC,if angle A=90^{circ},then (r_ | vedclass

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(D) In $	riangle ABC$,$\angle A=90^{\circ}$. The exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{

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$2 r_1^2$

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(D) In $	riangle ABC$,$\angle A=90^{\circ}$. The exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Since $\angle A=90^{\circ}$,we have $a^2 = b^2 + c^2$ and $\Delta = \frac{1}{2}bc$.Also,$s = \frac{a+b+c}{2}$,so $s-a = \frac{b+c-a}{2}$,$s-b = \frac{a+c-b}{2}$,and $s-c = \frac{a+b-c}{2}$.We know that $r_1 = s-a$,$r_2 = s-c$,and $r_3 = s-b$ is not correct; rather $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$.For a right-angled triangle at $A$,$r_1 = s-a$,$r_2 = s-c$,and $r_3 = s-b$ is incorrect. The correct relations are $r_1 = s-a$,$r_2 = s-c$,$r_3 = s-b$ is false. Actually,$r_1 = s-a$,$r_2 = s-c$,$r_3 = s-b$ is not standard.Using the identity $(r_2-r_1)(r_3-r_1) = 2r_2r_3$ is incorrect. The correct derivation leads to $(r_2-r_1)(r_3-r_1) = 2r_1^2$.

In $\triangle ABC$,if $\angle A=90^{\circ}$,then $(r_2-r_1)(r_3-r_1)=$

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