4. The bisector $A L$ and the median $B M$ of triangle $A B C$ intersect at point $X$. Line $C X$ intersects side $A B$ at point $Y$. Find the area of triangle $C Y L$, given that $\angle B A C=60^{\circ}$ and $A L=x$.
Answer: $\frac{x^{2}}{4 \sqrt{3}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-17.jpg?height=423&width=668&top_left_y=908&top_left_x=774)

Solution. Let $P$ and $Q$ be the midpoints of segments $A Y$ and $C L$. Then $M P$ and $M Q$ are the midlines of triangles $A C Y$ and $A C L$, from which $M P \| C Y$ and $M Q \| A L$. Therefore,

$$
\frac{A Y}{Y B}=2 \cdot \frac{P Y}{Y B}=2 \cdot \frac{M X}{X B}=2 \cdot \frac{L Q}{L B}=\frac{C L}{L B}
$$

From this, it follows that triangles $Y B L$ and $A B C$ are similar. Therefore, $A C \| Y L$ and

$$
\angle A L Y=\angle L A C=\frac{1}{2} \angle B A C=30^{\circ} .
$$

Thus, $A Y L$ is an isosceles triangle with base $x$ and base angles of $30^{\circ}$. We obtain

$$
S_{C Y L}=S_{A Y L}=\left(\frac{x}{2}\right)^{2} \cdot \operatorname{tg} 30^{\circ}=\frac{x^{2}}{4 \sqrt{3}}
$$