Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d7ec6e60d14a8fdbf750g-37.jpg?height=254&width=410&top_left_y=91&top_left_x=526)
Answer: 3.

![](https://cdn.mathpix.com/cropped/2024_05_06_d7ec6e60d14a8fdbf750g-37.jpg?height=505&width=493&top_left_y=432&top_left_x=480)

Fig. 5: to the solution of problem 9.7

Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals are bisected by their intersection point $L$, it is a parallelogram (in particular, $AC = DX$). Therefore, $DX \parallel AC$. Since $AC \parallel ED$ by the problem's condition, the points $X, D, E$ lie on the same line.

Since $AC \parallel EX$, then $\angle EAX = \angle CAX = \angle AXE$, i.e., triangle $AEX$ is isosceles, $EA = EX$. Then

$$
ED = EX - XD = EA - AC = 15 - 12 = 3
$$