58. Determine the measure of angle $\hat{A}$ of triangle $ABC$, if it is known that the bisector of this angle is perpendicular to the line passing through the point of intersection of the altitudes and the center of the circumscribed circle of this triangle.
58. Let $O$ be the center of the circumscribed circle, $H$ the intersection point of the altitudes of $\triangle ABC$. Since the line $OH$ is perpendicular to the bisector of angle $A$, it intersects sides $AB$ and $AC$ at points $K$ and $M$ such that $|AK|=|AM|$. Thus, $|AO|=|OB|$ and $\widehat{AOB}=2 \hat{C}$ (assuming angle $C$ is acute), $\widehat{OAK}=90^{\circ}-\hat{C}=\widehat{HAM}$. Therefore, $\triangle OAK=\triangle HAM$ and $|OA|=|HA|=R$ (where $R$ is the radius of the circumscribed circle). If $D$ is the foot of the perpendicular dropped from $O$ to $BC$, then $|OD|=\frac{1}{2}|AH|=\frac{R}{2}$. Consequently, $\cos \hat{A} = \cos \widehat{DOC} = \frac{1}{2}$ and $A=60^{\circ}$.