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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 −((1/2))^2 =y^2 =x(x+1) AE=(2/3)×AD cos ∠EAF=(x/(AE)) cos ∠EAF=(((x+1)^2 +AD^2 −((1/2))^2 )/(2(x+1)AD)) (((x+1)^2 +x^2 +2x)/(2(x+1)AD))=(x/(AE)) (((x+1)^2 +x^2 +2x)/(2(x+1)))=((3x)/2) x^2 −x−1=0 ⇒x=((1+(√5))/2) ✓

$${y}^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$${AD}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={y}^{\mathrm{2}} ={x}\left({x}+\mathrm{1}\right) \\ $$$${AE}=\frac{\mathrm{2}}{\mathrm{3}}×{AD} \\ $$$$\mathrm{cos}\:\angle{EAF}=\frac{{x}}{{AE}} \\ $$$$\mathrm{cos}\:\angle{EAF}=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{AD}^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}\left({x}+\mathrm{1}\right){AD}} \\ $$$$\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{2}\left({x}+\mathrm{1}\right){AD}}=\frac{{x}}{{AE}} \\ $$$$\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{x}}{\mathrm{2}\left({x}+\mathrm{1}\right)}=\frac{\mathrm{3}{x}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\checkmark \\ $$

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