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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

we use sin relation⇒ ((BD)/(sin((A/2))))=((AB)/(sin(∠BDA))) ((CD)/(sin((A/2))))=((AD)/(sin(180−∠BDA)))=((AC)/(sin(∠BDA))) ⇒((BD)/(CD))=((AB)/(AC))=(c/b) ⇒((BD)/(CD))+1=(c/b)+1⇔((BD+DC)/(CD))=((c+b)/b)⇔((BC)/(CD))=((c+b)/b)⇒((CD)/(BC))=(b/(b+c)) (2) we have ((AC)/(sin(B)))=((BC)/(sin(A)));((AD)/(sin(B)))=((BD)/(sin((A/2)))) ⇒((AC)/(AD))=((BC)/(BD)).((sin((A/2)))/(sin(A)))⇔(b/k)=((BC)/(BD)).(1/(2cos((A/2)))) ⇒k=((2bBDcos((A/2)))/(BC)) ((BD)/(BC))=((BC−CD)/(BC))=1−((CD)/(BC))=1−(b/(b+c))=(c/(b+c)) ⇒k=2b.(c/(b+c))cos((A/2))=((2bc)/(b+c))cos((A/2))

$${we}\:{use}\:{sin}\:{relation}\Rightarrow \\ $$$$\frac{{BD}}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)}=\frac{{AB}}{{sin}\left(\angle{BDA}\right)} \\ $$$$\frac{{CD}}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)}=\frac{{AD}}{{sin}\left(\mathrm{180}−\angle{BDA}\right)}=\frac{{AC}}{{sin}\left(\angle{BDA}\right)} \\ $$$$\Rightarrow\frac{{BD}}{{CD}}=\frac{{AB}}{{AC}}=\frac{{c}}{{b}} \\ $$$$\Rightarrow\frac{{BD}}{{CD}}+\mathrm{1}=\frac{{c}}{{b}}+\mathrm{1}\Leftrightarrow\frac{{BD}+{DC}}{{CD}}=\frac{{c}+{b}}{{b}}\Leftrightarrow\frac{{BC}}{{CD}}=\frac{{c}+{b}}{{b}}\Rightarrow\frac{{CD}}{{BC}}=\frac{{b}}{{b}+{c}} \\ $$$$\left(\mathrm{2}\right) \\ $$$${we}\:{have} \\ $$$$\frac{{AC}}{{sin}\left({B}\right)}=\frac{{BC}}{{sin}\left({A}\right)};\frac{{AD}}{{sin}\left({B}\right)}=\frac{{BD}}{{sin}\left(\frac{{A}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow\frac{{AC}}{{AD}}=\frac{{BC}}{{BD}}.\frac{{sin}\left(\frac{{A}}{\mathrm{2}}\right)}{{sin}\left({A}\right)}\Leftrightarrow\frac{{b}}{{k}}=\frac{{BC}}{{BD}}.\frac{\mathrm{1}}{\mathrm{2}{cos}\left(\frac{{A}}{\mathrm{2}}\right)}\: \\ $$$$\Rightarrow{k}=\frac{\mathrm{2}{bBDcos}\left(\frac{{A}}{\mathrm{2}}\right)}{{BC}} \\ $$$$\frac{{BD}}{{BC}}=\frac{{BC}−{CD}}{{BC}}=\mathrm{1}−\frac{{CD}}{{BC}}=\mathrm{1}−\frac{{b}}{{b}+{c}}=\frac{{c}}{{b}+{c}} \\ $$$$\Rightarrow{k}=\mathrm{2}{b}.\frac{{c}}{{b}+{c}}{cos}\left(\frac{{A}}{\mathrm{2}}\right)=\frac{\mathrm{2}{bc}}{{b}+{c}}{cos}\left(\frac{{A}}{\mathrm{2}}\right) \\ $$

Commented by TawaTawa last updated on 05/Feb/20

God bless you sir. I appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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

withe plesur miss

$${withe}\:{plesur}\:{miss}\: \\ $$

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