8.3. In triangle $A B C$, the median $B M$ is drawn. Sides $A B$ and $B C$ form angles of $100^{\circ}$ and $40^{\circ}$ with the median, respectively, and side $A B$ is equal to 1. Find the length of $B M$.
Answer: $B M=1 / 2$. Solution. Mark point $D$, symmetric to vertex $A$ with respect to point $M$. Quadrilateral $A B C D$ is a parallelogram because its diagonals are bisected at the point of intersection. Therefore, $\quad \angle B D C=\angle \mathrm{ABD}=100^{\circ} \quad$ and $\quad \angle$ $B C D=180^{\circ}-\angle D B C-\angle B D C=40^{\circ}$. Thus, triangle $B D$ is isosceles, which means $2 \cdot B M=B D=D C=A B=1 . \quad$ Hence, $B M=1 / 2$.(The result can also be obtained without considering the parallelogram, but instead by drawing a midline through point $M$ parallel to $B C$ and calculating the angles similarly).

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