Start by removing that line from B to D, it's distracting, but add lines from K and L to the centre of the big circle O.

Call the angle at D , then the angle AOL (the B side) will be , (for angles on the same arc, the angle at the centre will be twice the angle at the circumference).

That means the angle LON will be

Repeat the procedure for the other side of the figure and hence show that the angle KON is also

It follows that the triangles LON and KON are congruent in which case LN will be the same length as KN.

Since N is equidistant from L and K, it will lie on the perpendicular bisector of LK, which will be a diameter of the small circle.

The base angles of the triangles LON and KON will be , in which case the angle LNK will be

Finally, the angle at C is which is double the angle LNK. Putting this with the fact that N lies on a diameter, it follows that N is the centre of the small circle.