The height $BL$ of the rhombus $ABCD$, dropped to the side $AD$, intersects the diagonal $AC$ at point $E$. Find $AE$, if $BL=$ $8, AL: LD=3: 2$.

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$A E$ is the bisector of triangle $B A L$.

## Solution

Since $A E$ is the bisector of triangle $B A L$, then $B E: E L=A B: A L=A D: A L=5: 3$. From the right triangle, it is clear that $A L: B L=3: 4$, which means $A L=6$. Additionally, $E L=3 / 8 B L=3$. Therefore, $A E=$ $\sqrt{A L^{2}+E L^{2}}=3 \sqrt{5}$

![](https://cdn.mathpix.com/cropped/2024_05_06_a7c46f4bbe36942e4ae1g-31.jpg?height=435&width=834&top_left_y=383&top_left_x=612)

Answer

$3 \sqrt{5}$.