+0 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.
+1632 The intersection point of the three perpendicular bisectors of a triangle is the center of its circumcircle. In otherwords, ABC is inscribed in a circle with center O. Since angle AOB = 90 degrees, angle ACB = 90/2 = 45 degrees by Inscribed Angle Theorem.
Angle OMC and angle ONC are both 90 degrees by construction. Because OMCN is a quadrilateral, its interior angle sum is 360 degrees, so Angles OMC + MCN + CNO + MON = 360 degrees. Angle MCN = angle ACB = 45 degrees. By substitution, we get 90 + 45 + 90 + MON = 360 degrees, and angle MON = 135 degrees.
+1632 The intersection point of the three perpendicular bisectors of a triangle is the center of its circumcircle. In otherwords, ABC is inscribed in a circle with center O. Since angle AOB = 90 degrees, angle ACB = 90/2 = 45 degrees by Inscribed Angle Theorem.
Angle OMC and angle ONC are both 90 degrees by construction. Because OMCN is a quadrilateral, its interior angle sum is 360 degrees, so Angles OMC + MCN + CNO + MON = 360 degrees. Angle MCN = angle ACB = 45 degrees. By substitution, we get 90 + 45 + 90 + MON = 360 degrees, and angle MON = 135 degrees.
