In AB^Δ C : II_( a) ^( 2) =^? 4R ( r_( a) − r ) I : incircle center I_( a) : excircle center corresponding A R: circumcircle radius r: incircle radius r


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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

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