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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.

 Oct 30, 2023
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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}.]

 Oct 30, 2023

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