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Area of $\triangle AOB =\frac{1}{2}$
Area of $\triangle AMN =\frac{4}{9} \times \frac{1}{2}=\frac{2}{9}$
Equation of $A B$ is $x+y=1$
$OA=1, AM=\sec \left(45^{\circ}-\theta\right)$
$AN=\sec \left(45^{\circ}-\theta\right) \cos \theta$
$MN=\sec \left(45^{\circ}-\theta\right) \sin \theta$
$\operatorname{Ar}(\triangle AMN )=\frac{1}{2} \times \sec ^2\left(45^{\circ}-\theta\right) \sin \theta \cdot \cos \theta=\frac{2}{9}$
$\Rightarrow \tan \theta=2, \frac{1}{2}$
$\tan \theta=2 \text { is rejected }$
$\frac{ AN }{ NB }=\frac{\lambda}{1}=\cot \theta=2$