Have a pure geometric solution?

Use the fact that the orthocenter and circumcenter are isogonal conjugates. This implies that the median of the triangle passes through the circumcenter of the triangle. Now, if the circumcenter does not lie on $AB$ then this implies that either the triangle is isosceles (which is impossible since isosceles triangles have same median and angle bisector and altitude) or the triangle is degenerate. Since none of these cases can happen this must mean that the circumcenter is on $AB$ which implies $ABC$ is a right triangle. Let the foot of the altitude from $C$ be $D$ . Similarities tell us that $\angle DCB = \angle CAB = \frac{90 ^ \circ}{4} = 22.5 ^ \circ$ . Therefore, our answer is $\boxed{22.5 ^ \circ}$ .

Here we go:

Let $D, E, F$ respectively be foot of altitude, foot of angle bisector, foot of median - all from $C$ to $AB$ .

Drop a perpendicular from $E$ to $BC$ in the point $M$ and let $N$ be the intersection of $MD, AC$ .

Then $CDEM$ is cyclic. Hence,

$\angle CMD = \angle CED = \angle CAD$ .

It follows that triangle $\triangle CMN \approx \triangle CAB$ .

This implies that $CD$ is the median from $C$ in triangle $CMN$ , hence $AMEN$ is a parallelogram.

Finally, $AN \parallel ME$ and since $ME \bot BC$ , it follows that $AC \bot BC$ .

Hence, smallest angle is $\angle B = {22.5}^0$ .