3. Let $A, O, B$ be collinear points, $O \in (AB)$. From the same side of line $AB$, the rays $[OC$ and $[OD$ are drawn such that $m(\angle AOC) = 50^\circ$, $m(\angle COD) = 90^\circ$. If $[OX$ and $[OY$ are the bisectors of angles $\angle AOD$ and $\angle BOC$, respectively, determine the measure of angle $\angle XOY$.
3. It results that $m(\nless B O D)=40^{\circ}, m(\nless A O X)=70^{\circ}, m(\nless B O Y)=65^{\circ}$ $.4 \mathrm{p}$ $m(\nless X O Y)=180^{\circ}-70^{\circ}-65^{\circ}=45^{\circ}$