The (final) statement is not true.
Consider an acute isosceles triangle with vertex angle $A$. Because the figure is symmetric about the altitude through $A$, any circle through $B$ and $C$ that meets $AB$ and $AC$ will meet these edges at points $D$ and $E$ such that $BD$ and $CE$ have intersection $H$ somewhere on that altitude.
On one extreme, the smallest circle under consideration is the one with diameter $BC$, which in fact leads to orthocenter $H$ (since $\angle BDC$ and $\angle BEC$ subtend a semicircle); on the other, the largest circle is the circumcircle of $\triangle ABC$, for which we have $H=A$. Other circles lead to points in between the orthocenter and $A$, so the conditions do not "usually" cause $H$ to be the orthocenter.
A note about the acuteness condition: were $\triangle ABC$ a right isosceles triangle, the two "extreme" circles mentioned above would coincide, and we would indeed have $H=A$ lie at the orthocenter. Were the triangle obtuse, then it would lie entirely within the smallest circle (with diameter $BC$), allowing no intersection points $D$ and $E$ with the sides of the triangle. (Extended sides are another matter.)