+1364 In triangle $ABC$, $M$ is the midpoint of $\overline{BC}$, and $N$ is the midpoint of $\overline{AC}$. The perpendicular bisectors of $BC$ and $AC$ intersect at a point $O$ inside the triangle. If $\angle AOB = 90^\circ$, then find the measure of $\angle MON$, in degrees.
+1533 Since ∠AOB=90∘, we have [\angle BOM = \frac{1}{2}\angle ABC] and [\angle AON = \frac{1}{2}\angle BAC.] Since N is the midpoint of AC, we have [\angle ACN = 90^\circ - \frac{1}{2}\angle BAC.] Similarly, since M is the midpoint of BC, we have [\angle BCM = 90^\circ - \frac{1}{2}\angle ABC.] Therefore, [\angle MON = \angle BCM + \angle ACN = 180^\circ - \frac{1}{2}\angle ABC - \frac{1}{2}\angle BAC = \boxed{90^\circ + \frac{1}{2}\angle A}.]
