5. Given a parallelogram $A B C D$ with angle $\angle B$ equal to $60^{\circ}$. Point $O$ is the center of the circumscribed circle of triangle $A B C$. Line $B O$ intersects the bisector of the exterior angle $\angle D$ at point $E$. Find the ratio $\frac{B O}{O E}$.
Answer: $1 / 2$.

Solution. Let $P$ be the point of intersection of the circumcircle of triangle $ABC$ with line $CD$, distinct from $C$, and let $F$ be the point diametrically opposite to point $B$. Then $ABCP$ is an inscribed trapezoid. Therefore, it is isosceles and $\angle APD = \angle BAP = \angle ABC = 60^{\circ}$. Consequently,

$$
\angle PAD = \angle BAD - \angle BAP = 120^{\circ} - 60^{\circ} = 60^{\circ}.
$$

Thus, $AD = DP$. Since $OA = OP$ as well, lines $AP$ and $OD$ are perpendicular. Moreover, $DO$ is the bisector of angle $\angle ADC$. Therefore, $DO$ and $DE$ are perpendicular as the bisectors of the external and internal angles. Let $Q$ be the point of intersection of the diagonals of the parallelogram. Then $AQ = QC$ and, therefore, $OQ$ is the altitude of the isosceles triangle $AOC$ with an angle of $120^{\circ}$. Thus, $\angle OAC = 30^{\circ}$ and, therefore, $OQ = \frac{OA}{2}$. On the other hand, $OQ$ is the midline of triangle $BDF$. Therefore, $OQ = \frac{DF}{2}$ and $DF = OA = OF$. Since triangle $ODE$ is right-angled, $F$ is the midpoint of its hypotenuse. Thus, $OE = 2OF = 2OB$.