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Question Number 80706 by TawaTawa last updated on 05/Feb/20

Answered by mind is power last updated on 05/Feb/20

$$weusesinrelation\Rightarrow$$$$\frac{BD}{sin\left(\frac{A}{2}\right)}=\frac{AB}{sin\left(\mathrm{\angle }BDA\right)}$$$$\frac{CD}{sin\left(\frac{A}{2}\right)}=\frac{AD}{sin(180-\mathrm{\angle }BDA)}=\frac{AC}{sin\left(\mathrm{\angle }BDA\right)}$$$$\Rightarrow \frac{BD}{CD}=\frac{AB}{AC}=\frac{c}{b}$$$$\Rightarrow \frac{BD}{CD}+1=\frac{c}{b}+1\iff \frac{BD+DC}{CD}=\frac{c+b}{b}\iff \frac{BC}{CD}=\frac{c+b}{b}\Rightarrow \frac{CD}{BC}=\frac{b}{b+c}$$$$\left(2\right)$$$$wehave$$$$\frac{AC}{sin\left(B\right)}=\frac{BC}{sin\left(A\right)};\frac{AD}{sin\left(B\right)}=\frac{BD}{sin\left(\frac{A}{2}\right)}$$$$\Rightarrow \frac{AC}{AD}=\frac{BC}{BD}.\frac{sin\left(\frac{A}{2}\right)}{sin\left(A\right)}\iff \frac{b}{k}=\frac{BC}{BD}.\frac{1}{2cos\left(\frac{A}{2}\right)}$$$$\Rightarrow k=\frac{2bBDcos\left(\frac{A}{2}\right)}{BC}$$$$\frac{BD}{BC}=\frac{BC-CD}{BC}=1-\frac{CD}{BC}=1-\frac{b}{b+c}=\frac{c}{b+c}$$$$\Rightarrow k=2b.\frac{c}{b+c}cos\left(\frac{A}{2}\right)=\frac{2bc}{b+c}cos\left(\frac{A}{2}\right)$$

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

$$witheplesurmiss$$

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