2. On the side AC of triangle ABC with an angle of 120 degrees at vertex B, points D and E are marked such that $AD = AB$ and $CE = CB$. A perpendicular $DF$ is dropped from point D to the line BE. Find the ratio $BD / DF$.
Answer. 2. Solution. Let $\angle C A B=\alpha, \angle A C B=\beta$. Since $A D=A B$ and $C E=C B$, we have

$\angle D B E=\angle D B A+\angle E B C-\angle A B C=\left(180^{\circ}-\alpha\right) / 2+\left(180^{\circ}-\beta\right) / 2-120^{\circ}=60^{\circ}-(\alpha+\beta) / 2=30^{\circ}$.

Thus, in the right triangle $B F D$, the angle at vertex $B$ is $30^{\circ}$, from which $B D / D F=2$.