4. On the side $A C$ of triangle $A B C$, a point $E$ is chosen. The bisector $A L$ intersects the segment $B E$ at point $X$. It turns out that $A X=X E$ and $A L=B X$. What is the ratio of the angles $A$ and $B$ of the triangle? (S. Berlov)
Answer. 2. Solution. Let the line passing through point $E$ parallel to $A L$ intersect lines $B C$ and $B A$ at points $P$ and $Q$ respectively. From the similarity of triangles $A B L$ and $Q B P$, we have $P Q / A L = B E / B X = B E / A L$, from which it follows that $P Q = B E$. Due to the parallelism of lines $A L$ and $P Q$, we have $\angle A Q E = \angle B A X = \angle X A E = \angle A E Q$, from which it follows that $A E = A Q$. Moreover, from the equality $A X = X E$, it follows that $\angle A E B = \angle A E X = \angle X A E$, from which it follows that $\angle A E B = \angle A Q E$. Thus, triangles $A Q P$ and $A E B$ are equal by two sides and the angle between them. Therefore, $A P = A B$ and $\angle B A E = \angle P A Q = 2 \angle C B A$, from which we obtain the answer.