In-AB-C-II-a-2-4R-r-a-r-I-incircle-center-I-a-excircle-center-corresponding-A-R-circumcircle-radius-r-incircle-radiu

Question Number 190104 by mnjuly1970 last updated on 27/Mar/23

In A \overset{Δ}{B C} :

{II}_{a}^{2} \overset{?}{=} 4 R (r_{a} - r)

I : i n c i r c l e c e n t e r

I_{a} : e x c i r c l e c e n t e r c o r r e s p o n d i n g A

R : c i r c u m c i r c l e r a d i u s

r : i n c i r c l e r a d i u s

r_{a} : e x c i r c l e r a d i u s c o r r e s p o n d i n g A

Answered by mr W last updated on 27/Mar/23

Commented by mr W last updated on 27/Mar/23

r_{a} = \frac{r s}{s - a}

r_{a} - r = \frac{r s}{s - a} - r = \frac{r a}{s - a} = \frac{2 r a}{- a + b + c}

{I I}_{a} = \frac{r_{a} - r}{\sin \frac{A}{2}}

{({I I}_{a})}^{2} = \frac{(r_{a} - r) (r_{a} - r)}{\sin^{2} \frac{A}{2}}

{({I I}_{a})}^{2} = \frac{4 (r_{a} - r) r a}{(1 - \cos A) (- a + b + c)}

{({I I}_{a})}^{2} = \frac{4 (r_{a} - r) r a}{(1 - \frac{b^{2} + c^{2} - a^{2}}{2 b c}) (- a + b + c)}

{({I I}_{a})}^{2} = \frac{8 (r_{a} - r) r a b c}{(- a + b + c) (a + b - c) (a - b + c)}

{({I I}_{a})}^{2} = \frac{(r_{a} - r) Δ a b c}{Δ^{2}}

{({I I}_{a})}^{2} = \frac{(r_{a} - r) a b c}{Δ}

{({I I}_{a})}^{2} = 4 R (r_{a} - r) ✓

Commented by mnjuly1970 last updated on 28/Mar/23

e x c e l l e n t s i r W t h a n k s a l o t \dots .