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

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_( a) : excircle radius corresponding A

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{In}\:\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\:\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{II}_{\:{a}} ^{\:\mathrm{2}} \:\overset{?} {=}\:\mathrm{4}{R}\:\left(\:{r}_{\:{a}} \:−\:{r}\:\right) \\ $$$$ \\ $$$$\:\:\:\:\mathrm{I}\::\:{incircle}\:\:{center} \\ $$$$\:\:\:\mathrm{I}_{\:{a}} \::\:{excircle}\:{center}\:{corresponding}\:{A} \\ $$$$\:\:\:{R}:\:{circumcircle}\:{radius} \\ $$$$\:\:\:\:\:{r}:\:{incircle}\:{radius} \\ $$$$\:\:\:\:\:{r}_{\:{a}} \::\:{excircle}\:{radius}\:{corresponding}\:{A} \\ $$

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

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

r_a =((rs)/(s−a)) r_a −r=((rs)/(s−a))−r=((ra)/(s−a))=((2ra)/(−a+b+c)) II_a =((r_a −r)/(sin (A/2))) (II_a )^2 =(((r_a −r)(r_a −r))/(sin^2 (A/2))) (II_a )^2 =((4(r_a −r)ra)/((1−cos A)(−a+b+c))) (II_a )^2 =((4(r_a −r)ra)/((1−((b^2 +c^2 −a^2 )/(2bc)))(−a+b+c))) (II_a )^2 =((8(r_a −r)rabc)/((−a+b+c)(a+b−c)(a−b+c))) (II_a )^2 =(((r_a −r)Δabc)/Δ^2 ) (II_a )^2 =(((r_a −r)abc)/Δ) (II_a )^2 =4R(r_a −r) ✓

$${r}_{{a}} =\frac{{rs}}{{s}−{a}} \\ $$$${r}_{{a}} −{r}=\frac{{rs}}{{s}−{a}}−{r}=\frac{{ra}}{{s}−{a}}=\frac{\mathrm{2}{ra}}{−{a}+{b}+{c}} \\ $$$${II}_{{a}} =\frac{{r}_{{a}} −{r}}{\mathrm{sin}\:\frac{{A}}{\mathrm{2}}} \\ $$$$\left({II}_{{a}} \right)^{\mathrm{2}} =\frac{\left({r}_{{a}} −{r}\right)\left({r}_{{a}} −{r}\right)}{\mathrm{sin}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}}} \\ $$$$\left({II}_{{a}} \right)^{\mathrm{2}} =\frac{\mathrm{4}\left({r}_{{a}} −{r}\right){ra}}{\left(\mathrm{1}−\mathrm{cos}\:{A}\right)\left(−{a}+{b}+{c}\right)} \\ $$$$\left({II}_{{a}} \right)^{\mathrm{2}} =\frac{\mathrm{4}\left({r}_{{a}} −{r}\right){ra}}{\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\right)\left(−{a}+{b}+{c}\right)} \\ $$$$\left({II}_{{a}} \right)^{\mathrm{2}} =\frac{\mathrm{8}\left({r}_{{a}} −{r}\right){rabc}}{\left(−{a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)} \\ $$$$\left({II}_{{a}} \right)^{\mathrm{2}} =\frac{\left({r}_{{a}} −{r}\right)\Delta{abc}}{\Delta^{\mathrm{2}} } \\ $$$$\left({II}_{{a}} \right)^{\mathrm{2}} =\frac{\left({r}_{{a}} −{r}\right){abc}}{\Delta} \\ $$$$\left({II}_{{a}} \right)^{\mathrm{2}} =\mathrm{4}{R}\left({r}_{{a}} −{r}\right)\:\checkmark \\ $$

Commented by mnjuly1970 last updated on 28/Mar/23

excellent sir W thanks alot....

$$\:{excellent}\:{sir}\:\:{W}\:{thanks}\:{alot}.... \\ $$

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