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Question Number 43621 by ajfour last updated on 12/Sep/18

Commented by math1967 last updated on 13/Sep/18

AD=(√(x^2 −1))   ,AC=(√(x^2 +x^2 −1))=(√(2x^2 −1))  △ABD∼△CED ∴((ED)/x)=(1/(√(x^2 −1)))  ⇒ED=(x/(√(x^2 −1)))  Again ((AC)/(CD))=((AE)/(ED))    [CE is bisector]  ⇒((√(2x^2 −1))/x)=((√(x^2 −1))/x)  now I cannot find x   Is it any mistake?

$${AD}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:,{AC}=\sqrt{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{1}}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\bigtriangleup{ABD}\sim\bigtriangleup{CED}\:\therefore\frac{{ED}}{{x}}=\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\Rightarrow{ED}=\frac{{x}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${Again}\:\frac{{AC}}{{CD}}=\frac{{AE}}{{ED}}\:\:\:\:\left[{CE}\:{is}\:{bisector}\right] \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}}{{x}}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}} \\ $$$${now}\:{I}\:{cannot}\:{find}\:{x}\: \\ $$$${Is}\:{it}\:{any}\:{mistake}? \\ $$

Commented by MJS last updated on 13/Sep/18

no real solution I think

$$\mathrm{no}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{I}\:\mathrm{think} \\ $$

Answered by ajfour last updated on 13/Sep/18

x(tan 2θ−tan θ)=1 = xsin θ  ⇒ (((2sin θcos θ)/(2cos^2 θ−1))−((sin θ)/(cos θ)))=sin θ  if sin θ ≠ 0, then          .....

$${x}\left(\mathrm{tan}\:\mathrm{2}\theta−\mathrm{tan}\:\theta\right)=\mathrm{1}\:=\:{x}\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\left(\frac{\mathrm{2sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{2cos}\:^{\mathrm{2}} \theta−\mathrm{1}}−\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\right)=\mathrm{sin}\:\theta \\ $$$${if}\:\mathrm{sin}\:\theta\:\neq\:\mathrm{0},\:{then} \\ $$$$\:\:\:\:\:\:\:\:.....\:\: \\ $$

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