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Question-35167




Question Number 35167 by behi83417@gmail.com last updated on 16/May/18
Commented by behi83417@gmail.com last updated on 16/May/18
refer to Q#35115  AM=AN,sin(A/2)=((MN)/(2AN))⇒MN=2AN.sin(A/2)  AMON,is cyclic,so:  AO.MN=AM.ON+AN.OM⇒  2r.AN=AO.2AN.sin(A/2)⇒AO=(r/(sin(A/2)))  AD=AO+OD=(r/(sin(A/2)))+r=r(1+(1/(sin(A/2))))  ((DC)/(sin(A/2)))=((AC)/(sinB))⇒DC=(b/(sinB))sin(A/2)=2R.sin(A/2)  ⇒DC=BD=2R.sin(A/2)  ABDC is cyclic.so:  AD.BC=AB.DC+AC.BD⇒  a.r.(1+(1/(sin(A/2))))=c.2R.sin(A/2)+b.2R.sin(A/2)  ⇒r(1+sin(A/2))=2R((b+c)/a).sin^2 (A/2)⇒  ⇒r=2R.((b+c)/a).((sin^2 (A/2))/(1+sin(A/2)))   .■  note:  R=((abc)/(4S_(AB^△ C) )),sin(A/2)=(√(((p−b)(p−c))/(bc))),p=((a+b+c)/2)
$${refer}\:{to}\:{Q}#\mathrm{35115} \\ $$$${AM}={AN},{sin}\frac{{A}}{\mathrm{2}}=\frac{{MN}}{\mathrm{2}{AN}}\Rightarrow{MN}=\mathrm{2}{AN}.{sin}\frac{{A}}{\mathrm{2}} \\ $$$${AMON},{is}\:{cyclic},{so}: \\ $$$${AO}.{MN}={AM}.{ON}+{AN}.{OM}\Rightarrow \\ $$$$\mathrm{2}{r}.{AN}={AO}.\mathrm{2}{AN}.{sin}\frac{{A}}{\mathrm{2}}\Rightarrow{AO}=\frac{{r}}{{sin}\frac{{A}}{\mathrm{2}}} \\ $$$${AD}={AO}+{OD}=\frac{{r}}{{sin}\frac{{A}}{\mathrm{2}}}+{r}={r}\left(\mathrm{1}+\frac{\mathrm{1}}{{sin}\frac{{A}}{\mathrm{2}}}\right) \\ $$$$\frac{{DC}}{{sin}\frac{{A}}{\mathrm{2}}}=\frac{{AC}}{{sinB}}\Rightarrow{DC}=\frac{{b}}{{sinB}}{sin}\frac{{A}}{\mathrm{2}}=\mathrm{2}{R}.{sin}\frac{{A}}{\mathrm{2}} \\ $$$$\Rightarrow{DC}={BD}=\mathrm{2}{R}.{sin}\frac{{A}}{\mathrm{2}} \\ $$$${ABDC}\:{is}\:{cyclic}.{so}: \\ $$$${AD}.{BC}={AB}.{DC}+{AC}.{BD}\Rightarrow \\ $$$${a}.{r}.\left(\mathrm{1}+\frac{\mathrm{1}}{{sin}\frac{{A}}{\mathrm{2}}}\right)={c}.\mathrm{2}{R}.{sin}\frac{{A}}{\mathrm{2}}+{b}.\mathrm{2}{R}.{sin}\frac{{A}}{\mathrm{2}} \\ $$$$\Rightarrow{r}\left(\mathrm{1}+{sin}\frac{{A}}{\mathrm{2}}\right)=\mathrm{2}{R}\frac{{b}+{c}}{{a}}.{sin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}}\Rightarrow \\ $$$$\Rightarrow\boldsymbol{{r}}=\mathrm{2}\boldsymbol{{R}}.\frac{\boldsymbol{{b}}+\boldsymbol{{c}}}{\boldsymbol{{a}}}.\frac{\boldsymbol{{sin}}^{\mathrm{2}} \frac{\boldsymbol{{A}}}{\mathrm{2}}}{\mathrm{1}+\boldsymbol{{sin}}\frac{\boldsymbol{{A}}}{\mathrm{2}}}\:\:\:.\blacksquare \\ $$$${note}: \\ $$$${R}=\frac{{abc}}{\mathrm{4}{S}_{{A}\overset{\bigtriangleup} {{B}C}} },{sin}\frac{{A}}{\mathrm{2}}=\sqrt{\frac{\left({p}−{b}\right)\left({p}−{c}\right)}{{bc}}},{p}=\frac{{a}+{b}+{c}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 16/May/18
AOD may not be straight, Sir.
$${AOD}\:{may}\:{not}\:{be}\:{straight},\:{Sir}. \\ $$

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