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Question Number 201793 by sonukgindia last updated on 12/Dec/23

Answered by mr W last updated on 12/Dec/23

Commented by mr W last updated on 12/Dec/23

$$y^2={(x+1)}^2-1^2=x^2+2x$$$${AD}^2-{\left(\frac{1}{2}\right)}^2=y^2=x(x+1)$$$$AE=\frac{2}{3}\times AD$$$$\cos \mathrm{\angle }EAF=\frac{x}{AE}$$$$\cos \mathrm{\angle }EAF=\frac{{(x+1)}^2+{AD}^2-{\left(\frac{1}{2}\right)}^2}{2(x+1)AD}$$$$\frac{{(x+1)}^2+x^2+2x}{2(x+1)AD}=\frac{x}{AE}$$$$\frac{{(x+1)}^2+x^2+2x}{2(x+1)}=\frac{3x}{2}$$$$x^2-x-1=0$$$$\Rightarrow x=\frac{1+\sqrt{5}}{2}✓$$

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