+1557 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.
+42 Notice how the two perpendicular bisectors of a triangle meet at a point. This means point O is the circumcenter of the triangle (I do comp math not proofs plz forgive). By the inscribed angle theorem, BCA is 45 degrees (because point O is the center and BOA is 90 degrees). Because angles OMC and ONC are both equal to 90 degrees, quadrilateral OMCN is cyclic. Therefore, angles NOM and NCM must add up to 180 degrees, meaning angle MON is equal to 135 degrees.
Feel free to tell my if I made a mistake! :D
