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

Commented by math1967 last updated on 13/Sep/18

$$AD=\sqrt{x^2-1},AC=\sqrt{x^2+x^2-1}=\sqrt{2x^2-1}$$$$△ABD\sim △CED\therefore \frac{ED}{x}=\frac{1}{\sqrt{x^2-1}}$$$$\Rightarrow ED=\frac{x}{\sqrt{x^2-1}}$$$$Again\frac{AC}{CD}=\frac{AE}{ED}\left[CEisbisector\right]$$$$\Rightarrow \frac{\sqrt{2x^2-1}}{x}=\frac{\sqrt{x^2-1}}{x}$$$$nowIcannotfindx$$$$Isitanymistake?$$

Commented by MJS last updated on 13/Sep/18

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

Answered by ajfour last updated on 13/Sep/18

$$x(\tan 2\theta -\tan \theta )=1=x\sin \theta$$$$\Rightarrow (\frac{2\sin \theta \cos \theta }{2\cos {}^2\theta -1}-\frac{\sin \theta }{\cos \theta })=\sin \theta$$$$if\sin \theta \ne 0,then$$$$.....$$

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