Question 63
The extensions of the sides $AB$ and $CD$ of a convex quadrilateral $ABCD$ meet at point $P$ and the extensions of the sides $BC$ and $AD$ meet at point $Q$. The point $O$ from the interior of the quadrilateral is such that $\measuredangle BOP = \measuredangle DOQ$. Prove that $\measuredangle AOB + \measuredangle COD = 180^\circ$.