Question Number 80706 by TawaTawa last updated on 05/Feb/20
Answered by mind is power last updated on 05/Feb/20
$${we}\:{use}\:{sin}\:{relation}\Rightarrow \\ $$$$\frac{{BD}}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)}=\frac{{AB}}{{sin}\left(\angle{BDA}\right)} \\ $$$$\frac{{CD}}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)}=\frac{{AD}}{{sin}\left(\mathrm{180}−\angle{BDA}\right)}=\frac{{AC}}{{sin}\left(\angle{BDA}\right)} \\ $$$$\Rightarrow\frac{{BD}}{{CD}}=\frac{{AB}}{{AC}}=\frac{{c}}{{b}} \\ $$$$\Rightarrow\frac{{BD}}{{CD}}+\mathrm{1}=\frac{{c}}{{b}}+\mathrm{1}\Leftrightarrow\frac{{BD}+{DC}}{{CD}}=\frac{{c}+{b}}{{b}}\Leftrightarrow\frac{{BC}}{{CD}}=\frac{{c}+{b}}{{b}}\Rightarrow\frac{{CD}}{{BC}}=\frac{{b}}{{b}+{c}} \\ $$$$\left(\mathrm{2}\right) \\ $$$${we}\:{have} \\ $$$$\frac{{AC}}{{sin}\left({B}\right)}=\frac{{BC}}{{sin}\left({A}\right)};\frac{{AD}}{{sin}\left({B}\right)}=\frac{{BD}}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow\frac{{AC}}{{AD}}=\frac{{BC}}{{BD}}.\frac{{sin}\left(\frac{{A}}{\mathrm{2}}\right)}{{sin}\left({A}\right)}\Leftrightarrow\frac{{b}}{{k}}=\frac{{BC}}{{BD}}.\frac{\mathrm{1}}{\mathrm{2}{cos}\left(\frac{{A}}{\mathrm{2}}\right)}\: \\ $$$$\Rightarrow{k}=\frac{\mathrm{2}{bBDcos}\left(\frac{{A}}{\mathrm{2}}\right)}{{BC}} \\ $$$$\frac{{BD}}{{BC}}=\frac{{BC}−{CD}}{{BC}}=\mathrm{1}−\frac{{CD}}{{BC}}=\mathrm{1}−\frac{{b}}{{b}+{c}}=\frac{{c}}{{b}+{c}} \\ $$$$\Rightarrow{k}=\mathrm{2}{b}.\frac{{c}}{{b}+{c}}{cos}\left(\frac{{A}}{\mathrm{2}}\right)=\frac{\mathrm{2}{bc}}{{b}+{c}}{cos}\left(\frac{{A}}{\mathrm{2}}\right) \\ $$
Commented by TawaTawa last updated on 05/Feb/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Commented by mind is power last updated on 05/Feb/20
$${withe}\:{plesur}\:{miss}\: \\ $$