In a right angled ${\Delta }^{le}$ , the hypotenuse is $2\sqrt{2}$  times the length of perpendicular drawn from the opposite vertex to the hypotenuse, then the other two angles are

$\tan A=\frac{x}{AM}\Rightarrow AM=\frac{x}{\tan A}$

Similarly $CM=\frac{x}{\tan C}$

$AM+CM=2\sqrt{2}x$

$2\sqrt{2}x=\frac{x}{\tan A}+\frac{x}{\tan C}$

$2\sqrt{2}=\frac{1}{\tan A}+\frac{1}{\tan C}$

$\tan A+\tan C=2\sqrt{2}\tan A\tan C$

Also $A+C=\frac{\pi }{2}$

$\frac{\tan A+\tan C}{1-\tan A\tan C}=\infty$

$\tan A\tan C=1$

$\tan A+\tan C=2\sqrt{2}$

$\tan A\tan C=1$

By solving these two we get $\tan A=\sqrt{2}+1$

$\tan C=\sqrt{2}-1$

$C=\frac{\pi }{8}$  ,$A=\frac{3\pi }{8}$