If in a triangle ABC , (a - b) (s - c) = (b - c) (s - a), prove that $r_{1}, r_{2}, r_{3}$ are in A.P

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Solution

$r_{1}, r_{2}, r_{3}$ are in A.P., if $r_{2} - r_{1} = r_{3} - r_{2}$
or $\frac{S}{s - b} - \frac{S}{s - a} = \frac{S}{s - c} - \frac{S}{s - b}$
or $\frac{b - a}{s - b} = \frac{c - b}{s - c}$
or $(a - b) (s - c) = (b - c) (s - a)$
which is given

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Similar questions

\frac{a - b}{b - c}

\frac{s - a}{s - c},

r_{1}, r_{2}, r_{3}

\frac{a - b}{b - c} = \frac{s - a}{s - c},

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(\frac{b - c}{r_{1}}) + (\frac{c - a}{r_{2}}) + (\frac{a - b}{r_{3}}) = 0

\frac{r_{1}}{(s - b) (s - c)} + \frac{r_{2}}{(s - c) (s - a)} + \frac{r_{3}}{(s - a) (s - b)} = \frac{3}{r}

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