In triangle ABC,suppose the radius of the exc | vedclass

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(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area and $s$ is the semi-p

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$r$

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(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area and $s$ is the semi-perimeter of the triangle.Then,$ab - r_1 r_2 = ab - \frac{\Delta^2}{(s-a)(s-b)}$.Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $ab - r_1 r_2 = ab - s(s-c)$.Substituting $s = \frac{a+b+c}{2}$,we get $ab - \frac{a+b+c}{2} \cdot \frac{a+b-c}{2} = ab - \frac{(a+b)^2 - c^2}{4} = \frac{4ab - (a^2 + 2ab + b^2) + c^2}{4} = \frac{c^2 - (a-b)^2}{4} = \frac{c-(a-b)}{2} \cdot \frac{c+(a-b)}{2} = (s-a)(s-b)$.Now,$\frac{ab - r_1 r_2}{r_3} = \frac{(s-a)(s-b)}{\frac{\Delta}{s-c}} = \frac{(s-a)(s-b)(s-c)}{\Delta} = \frac{\Delta^2}{s \Delta} = \frac{\Delta}{s} = r$.

In $\triangle ABC$,suppose the radius of the excircle opposite to angle $A$ is denoted by $r_1$,similarly $r_2$ for angle $B$,and $r_3$ for angle $C$. If $r$ is the radius of the inscribed circle,then what is the value of $\frac{ab - r_1 r_2}{r_3}$?

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