Reflect the triangle $A B C$ over the sides $A B$ and $A C$, thus obtaining the triangles $B A C_{1}$ and $C A B_{1}$. Then,

$$
A C_{1}=A C=A B=A B_{1}, \angle A C_{1} B=\angle A C B=\angle A B C=\angle A B_{1} C=80^{\circ}
$$

and

$$
\angle C_{1} A B_{1}=3 \cdot 20^{\circ}=60^{\circ} ;
$$

therefore, the triangle $C_{1} A B_{1}$ is equilateral. Due to the reflection, the triangle $B C_{1} E$ is congruent to the triangle $B C E$, so $\angle B C_{1} E=\angle B C E=50^{\circ}$ and thus

$$
\angle A C_{1} E=\angle A C_{1} B-\angle B C_{1} E=80^{\circ}-50^{\circ}=30^{\circ} .
$$

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This implies that the line $C_{1} E$ is a symmetry axis of the equilateral triangle $A C_{1} B_{1}$. Let $F^{*}$ be the intersection point of the lines $C_{1} E$ and $A C$. The line $C_{1} E$ is also a symmetry axis of the triangle $A F^{*} B_{1}$, so $\angle F^{*} B_{1} A=\angle F^{*} A B_{1}=20^{\circ}$, hence $\angle F^{*} B_{1} C=\angle A B_{1} C-\angle F^{*} B_{1} A=60^{\circ}$.

Considering that the reflection of the triangle $F^{*} B C$ over the line $A C$ is the triangle $F^{*} B_{1} C$, $\angle F^{*} B C=60^{\circ} ;$ therefore, the points $F^{*}$ and $F$ coincide.

Since the line $B C$ is parallel to the line $B_{1} C_{1}$ due to the axial symmetry over the angle bisector of $\angle B A C$, the angle between the lines $E F$ and $B C$ is the same as the angle between the lines $E F$ (i.e., $C_{1} E$) and $B_{1} C_{1}$, which is half of the angle $\angle B_{1} C_{1} A$, i.e., $30^{\circ}$.
In triangle $A B C$, $A B=A C$ and $\angle B A C=20^{\circ}$. Let $E$ and $F$ be the points on sides $A B$ and $A C$ such that $\angle B C E=50^{\circ}$ and $\angle C B F=60^{\circ}$. What angle do the lines $B C$ and $E F$ form? (H)