If in a triangle $r_{1} = r_{2} + r_{3} + r,$ prove that the triangle is right angled.

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Solution

We have
$r_{1} = r_{2} + r_{3} + r$ or $r_{1} - r = r_{2} + r_{3}$
$\Rightarrow \frac{Δ}{s - a} - \frac{Δ}{s} = \frac{Δ}{s - b} + \frac{Δ}{s - c}$
$\Rightarrow = \frac{Δ a}{s (s - a)} = \frac{Δ (2 s - b - c)}{(s - b) (s - c)} = \frac{Δ a}{(s - b) (s - c)}$
$\Rightarrow s (s - a) = (s - b) (s - c)$
$\Rightarrow s^{2} - s a = s^{2} - (b + c) s + b c$
$\Rightarrow 2 s (b + c - a) = 2 b c$
$\Rightarrow = (a + b + c) (b + c - a) = 2 b c$
$\Rightarrow (b + c)^{2} - a^{2} = 2 b c$ $\Rightarrow b^{2} + c^{2} = a^{2}$
Hence, the triangle is right angled.

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