+1525 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.
+14995 Find the measure of angle MON, in degrees.
\(\alpha ,\beta ,\gamma\ be\ \angle \ of\ \triangle ABC\\\ \angle ACO\ is\ \angle 1\\ \angle BCO\ is \ \angle 2\)
\(\angle 1=\alpha -45°\\ \angle 2=\beta-45°\\ \angle AOB=90°\\ \gamma =\frac{1}{2}\cdot 90°=45°\ (Peripheral\ angle\ on\ the\ perimeter)\\ \gamma=\angle 1+ \angle 2=45°\\ \alpha +\beta=180°-\gamma=180°-45°=135° \)
\(\angle\ MON=(180°-90°-\angle 1)+(180°-90°-\angle 2)\\ \angle\ MON=180°-(\angle 1 +\angle 2)=180°-45°\\ \color{blue}\angle\ MON=135°\)
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