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.
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,
\[\angle MON = \angle DNM + \angle EOM = 2 \angle AON = \boxed{90^\circ}\].
Let's start by drawing a diagram to help us understand the problem better.
[DIAGRAM]
We are given that $M$ is the midpoint of $\overline{BC}$ and $N$ is the midpoint of $\overline{AC}$. Let the perpendicular bisectors of $BC$ and $AC$ intersect at a point $O$. We are also given that $\angle AOB = 90^\circ$.
Now, we need to find the measure of $\angle MON$. To do this, let's analyze the information we have.
Since $M$ is the midpoint of $\overline{BC}$, we know that $\overline{OM}$ is the perpendicular bisector of $\overline{BC}$.
Similarly, since $N$ is the midpoint of $\overline{AC}$, we know that $\overline{ON}$ is the perpendicular bisector of $\overline{AC}$.
The point where the perpendicular bisectors intersect, i.e., point $O$, is equidistant from the vertices $A$ and $C$. This means that $\overline{AN} \cong \overline{CN}$, and $\angle ANO = \angle CNO$.
Since $\overline{OM}$ is perpendicular to $\overline{BC}$, we know that $\angle MOB = 90^\circ$, and $\angle MOC = 90^\circ$.
Now, let's look at triangle $MON$.
$\angle MNO$ is the angle subtended by arc $\overline{MO}$ on the circumcircle of triangle $MON$.
Similarly, $\angle MON$ is the angle subtended by arc $\overline{MN}$ on the circumcircle of triangle $MON$.
Since $O$ is the circumcenter of triangle $MON$, we know that $\angle MNO = \angle ONO = \frac{1}{2} \angle MON$.
Therefore, $\angle MON = 2\angle MNO$.
To find the measure of $\angle MON$, we need to find the measure of $\angle MNO$.
From our previous analysis, we know that $\angle MNO = \frac{1}{2} \angle MON$.
Since $\angle MOB = 90^\circ$, we have $\angle MON = 2\angle MNO = 2(\frac{1}{2} \angle MON) = \angle MOB = 90^\circ$.
So, the measure of $\angle MON$ is $\boxed{90^\circ}$.
To find the measure of $\angle MON$, we need to determine the relationship between this angle and other angles in the triangle.
Let's start with $\angle MOC$ and $\angle NOC$. Since $O$ is the intersection point of the perpendicular bisectors of $BC$ and $AC$, it must lie on the circumcircle of triangle $ABC$. Thus, $\angle MOC$ and $\angle NOC$ are angles inscribed in the same arc $MC$ and $NC$ respectively. By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of its intercepted arc. Therefore, $\angle MOC = \frac{1}{2} \angle ABC$ and $\angle NOC = \frac{1}{2} \angle BAC$.
Next, consider $\angle BOC$. Since $\angle AOB = 90^\circ$, we can conclude that $AB$ is a diameter of the circumcircle of triangle $ABC$. By the Inscribed Angle Theorem, any angle inscribed in a semicircle is a right angle. Therefore, $\angle BOC = 90^\circ$.
Now we can find $\angle MON$. Notice that $\angle MON = \angle MOC + \angle NOC$. Using the relationships we found earlier, we have:
$\angle MON = \frac{1}{2} \angle ABC + \frac{1}{2} \angle BAC$
Since $\angle ABC$ and $\angle BAC$ are angles of triangle $ABC$, their measures can be determined using the angles in a triangle sum to 180 degrees.
$180^\circ = \angle ABC + \angle BAC + \angle ACB$
Since $M$ and $N$ are midpoints, we can conclude that $AM = MC$ and $BN = NC$. Therefore, triangle $ABC$ is an isosceles triangle.
By the Triangle Angle Sum Theorem, an isosceles triangle has two congruent angles. Thus, $\angle ABC = \angle BAC$.
Substituting this relationship into the equation above, we have:
$180^\circ = \angle ABC + \angle ABC + \angle ACB$
$180^\circ = 2\angle ABC + \angle ACB$
Since we know that $\angle BOC = 90^\circ$, we can conclude that $\angle ACB$ is the supplement of $\angle BOC$.
$\angle ACB + \angle BOC = 180^\circ$
Substituting this relationship again, we have:
$2\angle ABC + \angle BOC = 180^\circ$
Solving for $\angle ABC$, we find:
$\angle ABC = \frac{180^\circ - \angle BOC}{2}$
Substituting this into our expression for $\angle MON$, we get:
$\angle MON = \frac{1}{2} \left(\frac{180^\circ - \angle BOC}{2}\right) + \frac{1}{2} \angle BOC$
Simplifying further, we have:
$\angle MON = \frac{1}{4} (180^\circ - \angle BOC) + \frac{1}{2} \angle BOC$
$\angle MON = \frac{1}{4} \cdot 180^\circ - \frac{1}{4} \cdot \angle BOC + \frac{1}{2} \cdot \angle BOC$
$\angle MON = 45^\circ - \frac{1}{4} \cdot \angle BOC + \frac{1}{2} \cdot \angle BOC$
$\angle MON = 45^\circ + \frac{1}{4} \cdot \angle BOC$
Thus, the measure of $\angle MON$ is $\boxed{45}$ degrees.