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.

Answer 1

Let $D$ and $E$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively. Then quadrilateral $BCDE$ is a parallelogram, so $DE = BC$. Since $ANO$ is an isosceles right triangle with hypotenuse $AN$ (because $\angle AOB = 90^\circ$), $DN = NA = DO$.

Since $\angle DOM$ and $\angle DNM$ subtend the same arc, $\angle DOM = \angle DNM = \angle AON$. Similarly, $\angle EOM = \angle EON = \angle AON$. Therefore, by vertical angles,
$$\mathrm{\angle }MON=\mathrm{\angle }DNM+\mathrm{\angle }EOM=2\mathrm{\angle }AON=\boxed{{\displaystyle {90}^{\circ }}}$$.