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Question Number 30455 by ajfour last updated on 22/Feb/18

Commented by mrW2 last updated on 25/Feb/18

$$p=\frac{ac(a^2-b^2+c^2)}{a(a^2-b^2+c^2)+c(a^2+b^2-c^2)}$$$$q=\frac{c^2(a^2+b^2-c^2)}{a(a^2-b^2+c^2)+c(a^2+b^2-c^2)}$$$$\frac{p}{q}=\frac{a(a^2-b^2+c^2)}{c(a^2+b^2-c^2)}$$$$pleaseconfirm.$$

Commented by ajfour last updated on 25/Feb/18

$$yessir,withsomeeffortiobtained$$$$thesame.$$

Commented by ajfour last updated on 25/Feb/18

Commented by ajfour last updated on 25/Feb/18

$$\lambda \overline{a}+\mu (\overline{c}-\lambda \overline{a})=\eta (p\overline{c}-\overline{a})+\overline{a}$$$$ifBAisdividedintheratio$$$$p/(1-p).$$$$comparingcoefficientof\overline{c},$$$$\Rightarrow \mu =\eta p...\left(i\right)$$$$furthersince\mathrm{\angle }Bisbisected,$$$$\mu =\frac{\lambda a}{c+\lambda a},\eta =\frac{a}{a+pc}..\left(ii\right)$$$$and\mathit{\lambda }=\frac{\mathit{c}\cos \mathit{B}}{\mathit{a}}....\left(iii\right)$$$$from\left(i\right)and\left(ii\right):$$$$p=\frac{\mu }{\eta }=\frac{\lambda (a+pc)}{c+\lambda a}$$$$\Rightarrow p(c+\lambda a)=\lambda a+\lambda cp$$$$p=\frac{\lambda a}{\lambda a+c-\lambda c}now,using\left(iii\right)$$$$\mathit{p}=\frac{\mathit{c}\cos \mathit{B}}{\mathit{c}\cos \mathit{B}+\mathit{c}-\frac{{\mathit{c}}^2\cos \mathit{B}}{\mathit{a}}}$$$$=\frac{\mathit{a}\mathit{c}\cos \mathit{B}}{\mathit{a}\mathit{c}\cos \mathit{B}+\mathit{a}\mathit{c}-{\mathit{c}}^2\cos \mathit{B}}$$$$=\frac{(a^2+c^2-b^2)}{(a^2+c^2-b^2)+2ac-\frac{c}{a}(a^2+c^2-b^2)}$$$$pc=\frac{ac(a^2+c^2-b^2)}{a(a^2+c^2-b^2)+c(a^2+b^2-c^2)}$$$$qc=c-pc$$$$qc=\frac{c^2(a^2+b^2-c^2)}{a(a^2+c^2-b^2)+c(a^2+b^2-c^2)}.$$

Commented by mrW2 last updated on 25/Feb/18

$$\mathbb{GREAT}!$$

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