Geometry

Let  $BC = a, AC = b$ just to simplify our eyes a bit. 

Let's note that through the Pythagorean Theorem, we can find AM and BN in terms of a and b. 

We get $\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}$

and $\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}$

Because AM and BN both equal 1, we get the equations

$4b^2 + a^2 = 1^2 \\4a^2 + b^2 = 1^2$

Adding the two equations together and dividing, we get

$a^2 + b^2 = \dfrac{1+ 1}{5} = 2/5$

Since the hypotenuse is just the square root of a^2 + b^2, we get

$\sqrt{a^2+ b^2} = \sqrt{2/5}$

Thanks! :)