# Thread: Incricles and Excircles

1. ## Incricles and Excircles

If I have r as an inradius of a triangle and ra, rb, rc are the three radii of the 3 excircles of that triangle how can I show that:

1/ra+1/rb+1/rc=1/r?

2. Originally Posted by TheNcredibleHulk
If I have r as an inradius of a triangle and ra, rb, rc are the three radii of the 3 excircles of that triangle how can I show that:

1/ra+1/rb+1/rc=1/r?
Start here, with the formulas $\displaystyle r = A/s$, $\displaystyle r_a = A/(s-a)$, $\displaystyle r_b = A/(s-b)$, $\displaystyle r_c = A/(s-c)$, where A is the area of the triangle and $\displaystyle s = \tfrac12(a+b+c)$.

3. Im still not getting this.

4. Originally Posted by Opalg
Start here, with the formulas $\displaystyle r = A/s$, $\displaystyle r_a = A/(s-a)$, $\displaystyle r_b = A/(s-b)$, $\displaystyle r_c = A/(s-c)$, where A is the area of the triangle and $\displaystyle s = \tfrac12(a+b+c)$.
If you are willing to accept those formulas, then the calculation goes like this:

$\displaystyle \frac1{r_a}\, +\, \frac1{r_b}\, +\, \frac1{r_c} = \frac{(s-a)+(s-b)+(s-c)}A = \frac{3s-(a+b+c)}A = \frac{3s-2s}A = \frac sA = \frac1r.$

### 2/ha=1/r-1/ra=1/rb 1/rc

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