12.146. Through the vertices of an equilateral triangle $A B C$, parallel lines $A D, B E$ and $C F$ are drawn. The line $B E$ lies between the lines $A D$ and $C F$ and divides the distance between them in the ratio $m: n$, measured from the line $A D$. Find the angle $B C F$.
## Solution.

Let $A K$ be the common perpendicular of the given parallel lines, $L$ be the intersection point of $B E$ and $A K, A L: L K=m: n, B F \perp F K$ (Fig. 12.19).

Let $B C=a, \angle B C F=\alpha$.

In $\triangle B F C\left(\angle B F C=90^{\circ}\right): B F=a \sin \alpha, K L=B F=a \sin \alpha$,

$$
\begin{aligned}
& \angle C B L=\angle B C F=\alpha \\
& \angle A B L=60^{\circ}-\alpha, \text { and from } \triangle A L B\left(\angle A L B=90^{\circ}\right): A L=a \sin \left(60^{\circ}-\alpha\right) \\
& \text { Then } \frac{a \sin \left(60^{\circ}-\alpha\right)}{a \sin \alpha}=\frac{m}{n} ; \frac{\sqrt{3}}{2} \operatorname{ctg} \alpha-\frac{1}{2}=\frac{m}{n} \\
& \operatorname{ctg} \alpha=\frac{2 m+n}{n \sqrt{3}} ; \alpha=\operatorname{arctg} \frac{n \sqrt{3}}{2 m+n} \\
& \text { Answer: } \operatorname{arctg} \frac{n \sqrt{3}}{2 m+n}
\end{aligned}
$$