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Question Number 70163 by A8;15: last updated on 01/Oct/19

Answered by ajfour last updated on 01/Oct/19

Commented by ajfour last updated on 02/Oct/19

$$\frac{AB}{BC}=Q=\frac{x+y}{\sqrt{y^2+z^2}}$$$$x+y+e+d=2c-a....\left(i\right)$$$$\frac{x}{y}=\frac{a}{b}\mathrm{\&}\frac{\mathit{e}}{d}=\frac{b}{c}...\left(ii\right)$$$${(2c-a)}^2=4ac$$$$\Rightarrow 4{\left(\frac{c}{a}\right)}^2-8\left(\frac{c}{a}\right)+1=0$$$$\Rightarrow \frac{c}{a}=\lambda =\frac{8+\sqrt{64-16}}{8}=1+\frac{\sqrt{3}}{2}$$$${(\mathit{e}+d)}^2=4bc$$$$d(\frac{b}{c}+1)=2\sqrt{bc}$$$$\Rightarrow d=\frac{2c\sqrt{bc}}{b+c},\mathit{e}=\frac{2b\sqrt{bc}}{b+c}$$$$\frac{a-z}{z-b}=\frac{a}{b}\Rightarrow ab-bz=az-ab$$$$\Rightarrow z=\frac{2ab}{a+b}$$$$y^2=b^2-{(z-b)}^2$$$$=2bz-z^2=\frac{4{ab}^2(a+b)-4a^2{b}^2}{{(a+b)}^2}$$$$y=\frac{2b\sqrt{ab}}{a+b};x=\frac{2a\sqrt{ab}}{a+b}$$$$Q=\frac{AB}{BC}=\frac{x+y}{\sqrt{y^2+z^2}}$$$$=\frac{2\sqrt{ab}}{\sqrt{{\left(\frac{2b\sqrt{ab}}{a+b}\right)}^2+{\left(\frac{2ab}{a+b}\right)}^2}}$$$$=\frac{2(a+b)\sqrt{ab}}{\sqrt{4{ab}^2(b+a)}}=\sqrt{1+\frac{a}{b}}$$$$Nowtakingup\left(i\right)$$$$x+y+e+d=2c-a$$$$2\sqrt{ab}+2\sqrt{bc}=2c-a$$$$\Rightarrow 2\sqrt{\frac{b}{a}}+2\sqrt{\lambda \left(\frac{b}{a}\right)}=2\lambda -1$$$$\Rightarrow 2\sqrt{\frac{b}{a}}(1+\sqrt{\lambda })=2\lambda -1$$$$\sqrt{\frac{b}{a}}=\frac{2\lambda -1}{2(1+\sqrt{\lambda })}$$$$Q=\frac{AB}{BC}=\sqrt{1+\frac{4{(1+\sqrt{\lambda })}^2}{{(2\lambda -1)}^2}}$$$$\lambda =1+\frac{\sqrt{3}}{2}.$$

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