If $(r_{2} - r_{1}) (r_{3} - r_{1}) = 2 r_{2} r_{3}$ show that $A = 90^{\circ} .$

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Detailed Solution

$(r_{2} - r_{1}) (r_{3} - r_{1}) = 2 r_{2} r_{3}$

$\Rightarrow (\frac{Δ}{s - b} - \frac{Δ}{s - a}) (\frac{Δ}{s - c} - \frac{Δ}{s - a}) = 2 \frac{Δ}{s - b} \frac{Δ}{s - c}$

$\Rightarrow Δ (\frac{1}{s - b} - \frac{1}{s - a}) Δ (\frac{1}{s - c} - \frac{1}{s - a}) = \frac{2 Δ^{2}}{(s - b) (s - c)}$

$\Rightarrow (\frac{s - a - s + b}{(s - b) (s - a)}) (\frac{s - a - s + c}{(s - c) (s - a)}) = \frac{2}{(s - b) (s - c)}$

$\Rightarrow (b - a) (c - a) = 2 (s - a)^{2}$

$= 2 {(\frac{a + b + c}{2} - a)}^{2} = 2 {(\frac{b + c - a}{2})}^{2}$

$\Rightarrow b c - a b - a c + a^{2}$

$= \frac{2 (b^{2} + c^{2} + a^{2} + 2 b c - 2 c a - 2 a b)}{4}$

$\Rightarrow 2 b c - 2 a b - 2 a c + 2 a^{2} = b^{2} + c^{2} + a^{2} + 2 b c - 2 c a - 2 a b$

$2 a^{2} = b^{2} + c^{2} + a^{2} \Rightarrow a^{2} = b^{2} + c^{2}$

$\Rightarrow ⌊A = 90^{\circ}$

$△ A B C$ is a right angle triangle and $A = 90^{\circ} .$

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