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Question Number 113505 by Lekhraj last updated on 13/Sep/20

Answered by mr W last updated on 13/Sep/20

Commented by mr W last updated on 13/Sep/20

$$let\lambda =\frac{y}{x}$$$$\frac{CG}{BE}=\frac{x}{x+y}=\frac{1}{1+\frac{y}{x}}=\frac{1}{1+\lambda }$$$$CG=\frac{k}{1+\lambda }$$$$\frac{CF}{AF}=\frac{CG}{AE}=\frac{k}{2k(1+\lambda )}=\frac{1}{2(1+\lambda )}$$$$\frac{CF}{CA}=\frac{CF}{CF+AF}=\frac{1}{1+\frac{AF}{CF}}=\frac{1}{1+2(1+\lambda )}=\frac{1}{3+2\lambda }$$$$\frac{\mathrm{\Delta }BCF}{\mathrm{\Delta }BCA}=\frac{CF}{CA}=\frac{1}{3+2\lambda }$$$$\frac{\mathrm{\Delta }BCF}{\mathrm{\Delta }CDF}=\frac{y}{x}=\lambda$$$$\frac{green}{red}=\frac{\mathrm{\Delta }BCA}{\mathrm{\Delta }CDF}=\frac{\mathrm{\Delta }BCA}{\mathrm{\Delta }BCF}\times \frac{\mathrm{\Delta }BCF}{\mathrm{\Delta }CDF}=1$$$$\Rightarrow (3+2\lambda )\lambda =1$$$$\Rightarrow 2{\lambda }^2+3\lambda -1=0$$$$\Rightarrow \lambda =\frac{\sqrt{17}-3}{4}=\frac{y}{x}$$$$\Rightarrow \frac{x}{y}=\frac{4}{\sqrt{17}-3}=\frac{\sqrt{17}+3}{2}\approx 3.562$$

Commented by Lekhraj last updated on 14/Sep/20

$$T\mathrm{hanks}\mathrm{a}\mathrm{lot}\mathrm{sir}\mathrm{for}\mathrm{such}\mathrm{a}\mathrm{nice}\mathrm{and}$$$$\mathrm{great}\mathrm{solution}.\mathrm{Have}\mathrm{you}\mathrm{any}$$$$\mathrm{twitter}\mathrm{handle}\mathrm{sir}\mathrm{I}liketofollow$$$$you.$$

Commented by mr W last updated on 14/Sep/20

$$nosir.$$$$thanksforputtingquestionshere!$$

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