Question Number 27959 by ajfour last updated on 17/Jan/18
Commented by ajfour last updated on 17/Jan/18
$${Inspired}\:{with}\:{Q}.#\mathrm{27942} \\ $$
Commented by beh.i83417@gmail.com last updated on 17/Jan/18
$$\frac{\boldsymbol{{r}}}{\boldsymbol{{a}}/\mathrm{2}}=\frac{\boldsymbol{{A}}{x}}{{xC}}\Rightarrow\frac{\mathrm{2}{r}}{{a}}=\frac{\boldsymbol{{A}}{x}}{{xC}}\Rightarrow\frac{\mathrm{2}{r}+{a}}{{a}}=\frac{{b}}{{Cx}} \\ $$$${cosC}=\frac{{a}/\mathrm{2}}{{xC}}\Rightarrow{xC}=\frac{{a}}{\mathrm{2}{cosC}} \\ $$$$\Rightarrow\mathrm{2}{r}/{a}+\mathrm{1}=\frac{{b}}{\frac{{a}}{\mathrm{2}{cosC}}}=\frac{\mathrm{2}{bcosC}}{{a}}\Rightarrow \\ $$$$\mathrm{2}{r}+{a}=\mathrm{2}{bcosC}\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{bcosC}−{a}\right) \\ $$$$\left.\mathrm{1}\right){a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{bc} \\ $$$$\left.\mathrm{2}\right){cosC}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{bc}}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{bc}+{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{bc}}= \\ $$$$=\frac{\mathrm{2}{b}^{\mathrm{2}} −{bc}}{\mathrm{2}{bc}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{b}.\frac{\mathrm{2}{b}^{\mathrm{2}} −{bc}}{\mathrm{2}{bc}}−{a}\right)=\frac{\mathrm{2}{b}^{\mathrm{2}} −{bc}−{ac}}{\mathrm{2}{c}} \\ $$$$\Rightarrow\boldsymbol{{r}}=\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{c}}}−\frac{\boldsymbol{{a}}+\boldsymbol{{b}}}{\mathrm{2}}\:.\blacksquare \\ $$
Commented by ajfour last updated on 17/Jan/18
$${Sir},\:{what}\:{does}\:{Ax},\:{xC},\:..{mean}\:? \\ $$
Commented by beh.i83417@gmail.com last updated on 17/Jan/18
$${x},\:{is}\:{the}\:{intersection}\:{of}\:\boldsymbol{{AC}}\:{with}\: \\ $$$${perpendicular}\:{line}\:{drawn}\:{from}\:{midpoint} \\ $$$${of}\:\boldsymbol{{BC}}. \\ $$
Commented by beh.i83417@gmail.com last updated on 18/Jan/18
$${sir}\:{Ajfour}!\:{what}\:{about}\:{Q}#\mathrm{27942}\:? \\ $$
Commented by ajfour last updated on 18/Jan/18
$${thank}\:{you}\:{sir}. \\ $$