How was this geometry problem created? 
This is a standard High School Olympiad problem and for an experienced problem solver a quite easy solve. But how was this problem created. To pose a problem, I believe is much harder, than to solve a posed problem. 
Here the problem poser had to first make the figure up and then simultaneously realise that $ND$ had the wonderful property of being equal in magnitude to the circumradius. Is there a nifty way to find out these wonderful geometric properties? 
 A: Using GeoGebra is a fantastic way to come up with geometry problems such as this. Just screwing around with GeoGebra can give you very interesting and contest-able geometry problems, and I'd bet that this problem was created with GeoGebra or an equivalent program.
A: Let me show you another solution ;)... Maybe this is how they came up with this problem. Manipulating midpoints, orthogonality and parallel lines. 
Draw the perpendicular of edge $AC$ at point $A$ and let it intersect the line $BN$ at point $P$. Since $\angle \, CBP = 90^{\circ} = \angle \, CAP$, it follows that quadrilateral $BCAP$ is inscribed in a circle (the circumcircle of triangle $ABC$) and $PC$ is a diameter of that circle. Since line $FK \equiv NK$ is perpendicular to $AC$, it is parallel to $AP$. As $FK$ passes trough the midpoint $F$ of segment $AB$ and is parallel to $AP$, its intersection point $N$ with segment $BP$ is the midpoint of $BP$ (i.e. $FN$ is the midsegment of triangle $BAP$). Consequently, in triangle $BCP$ the points $N$ and $D$ are midpoints of edges $BP$ and $BC$ respectively, making $ND$ the midsegmetn of triangle $BCP$ parallel to the diameter $CP$ and half of its length, i.e. $ND = \frac{1}{2} CP$ equals the radius of the circumcirlce of triangle $ABC$. 
A: This is NOT an answer to your question but is a shorter alternative version to the provided solution.
It should be clear that why the square-marked angles are right angles. Then, we can say that FNBD, and FBDO are cyclic quadrilaterals.

This means B, D, O, F, and N are all on the red dotted circle. Then, $\angle BNO = 90^0$. 
Result follows from the fact that DN = OB because they are the diagonals of the rectangle BDON. 
