5. Let $O$ be the point of intersection of the medians of triangle $ABC$. Find the length of the median drawn from vertex $A$, if $\angle BAC=35^{\circ}, \angle BOC=145^{\circ}, BC=a$.
Solution. Let the length of the desired median $AD$ be $m$. On the line $AD$ outside the triangle, mark a point $F$ such that $OD = DF = m / 3$. The quadrilateral $OBFC$ is a parallelogram because its diagonals are bisected by their point of intersection (by the condition $BD = DC$, and $OD = DF$ by construction). In a parallelogram, opposite angles are equal, hence

$$
\angle CFB = \angle BOC.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_7f07b77cc782d9cbfea6g-3.jpg?height=368&width=377&top_left_y=244&top_left_x=1545)

In the quadrilateral $ABFC$, by the condition and due to the equality (1), the sum of the opposite angles $BAC$ and $CFB$ is $180^{\circ}$. Therefore, a circle can be circumscribed around the quadrilateral $ABFC$. It is known that if two chords of a circle, $BC$ and $AF$, intersect at point $D$, then the product of the segments of one chord equals the product of the segments of the other chord, i.e., $BD \cdot DC = AD \cdot DF \Leftrightarrow (a / 2)^2 = m \cdot m / 3$. From this, $m = a \sqrt{3} / 2$.

Answer: $\frac{a \sqrt{3}}{2}$.