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I want to prove that for a triangle by side $a, b, c $, Inradius is:$$r=\dfrac{S}{P}=\frac{1}{2}\sqrt{\dfrac{(a+b-c)(a+c-b)(b+c-a)}{a+b+c}}$$. Such that $S$ is area and $P$ is semiperimeter. but I cant prove it. Thank you.http://mathworld.wolfram.com/Inradius.html

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  • $\begingroup$ Do you know Heron's formula? en.wikipedia.org/wiki/Heron%27s_formula $\endgroup$ – Robert Z Aug 19 '16 at 10:14
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For the first part i.e. $r = \frac{S}{P}$,

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Since $S = \frac{1}{2} a r + \frac{1}{2} b r + \frac{1}{2} c r = P r $,

then the equation holds.

For the second parts $\frac{S}{P}= \frac 12\sqrt{\frac{(b+c-a)(c+a-b)(a+b-c)}{a+b+c}}$,

by Heron's formula, $S = \sqrt{P (P-a)(P-b) (P-c)}$.

Substitute into the formula $\frac{S}{P}$ then using the definition of $P$, one will get the answer.

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Let $I$ be the incenter of $ABC$ and $r$ be the inradius.
Since $I$ always lies in the interior of $ABC$ (it is the common point to angle bisectors) we have:

$$ 2[ABC]=2[IAB]+2[IAC]+2[IBC]= rc+rb+ra = r(a+b+c), $$ so $$ r=\frac{2[ABC]}{a+b+c}. $$ If we compute $[ABC]$ through Heron's formula, $$ r = \frac{1}{2}\sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c)}{a+b+c}} $$ readily follows.

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