If triangle ABC is right-angled at A,then r_2 | vedclass

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(A) In a triangle $ABC$,the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,and the inradius

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$r_1-r$

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(A) In a triangle $ABC$,the exradii are given by $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,and the inradius is $r = \frac{\Delta}{s}$.Given $\angle A = 90^{\circ}$,we have $a^2 = b^2 + c^2$.We know the identity $r_1+r_2+r_3-r = 4R$.Also,for a right-angled triangle at $A$,$r_1 = s 	an(A/2) = s 	an(45^{\circ}) = s$.Since $s = \frac{a+b+c}{2}$,we have $r_1 = \frac{a+b+c}{2}$.Also,$r = \frac{\Delta}{s} = \frac{bc/2}{(a+b+c)/2} = \frac{bc}{a+b+c}$.Using the property $r_2+r_3 = 4R \cos^2(A/2) \sin(B/2) \cos(C/2) + \dots$ is complex,so we use $r_1-r = \frac{\Delta}{s-a} - \frac{\Delta}{s} = \frac{\Delta a}{s(s-a)}$.For $\angle A = 90^{\circ}$,$r_2+r_3 = 2R \sin(A) = 2R(1) = 2R$.However,using $r_1-r = 4R \sin(A/2) \cos(B/2) \cos(C/2) = 4R \sin(45^{\circ}) \cos(B/2) \cos(C/2) = 2R \sqrt{2} \cos(B/2) \cos(C/2)$.Actually,for $\angle A = 90^{\circ}$,$r_1 = s$ and $r_2+r_3 = 2R+r-r_1$ is not standard. The correct relation is $r_2+r_3 = r_1-r$.

If $\triangle ABC$ is right-angled at $A$,then $r_2+r_3$ is equal to

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