4. In a right-angled triangle $A B C$ with a right angle at $B$, the bisector $B L$ and the median $C M$ are drawn, intersecting at point $D$. The line $A D$ intersects side $B C$ at point $E$. Find the area of triangle $A E L$, given that $E L=x$.
Answer: $\frac{x^{2}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_bd69057b1c245517edf6g-20.jpg?height=431&width=580&top_left_y=710&top_left_x=812)

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

$$
\frac{A L}{L C}=2 \cdot \frac{P L}{L C}=2 \cdot \frac{M D}{D C}=2 \cdot \frac{Q E}{E C}=\frac{B E}{E C}
$$

Then triangles $L C E$ and $A C B$ are similar, and thus angle $L E C$ is a right angle. Moreover, $\angle L B E=45^{\circ}$. Therefore, $B E L$ is an isosceles right triangle with leg $x$. We obtain

$$
S_{A E L}=S_{B E L}=\frac{x^{2}}{2}
$$