In triangle $ABC$, the height $BD$ is 6, the median $CE$ is 5, and the distance from point $K$ of intersection of segments $BD$ and $CE$ to side $AC$ is 1.

Find the side $AB$.
Continue the median $C E$ until it intersects with the line passing through vertex $B$ parallel to $A C$, and consider the similar triangles formed.

## Solution

Let $P$ be the point of intersection of the line $C E$ with the line passing through vertex $B$ parallel to $A C$. From the equality of triangles $P B E$ and $C A E$, it follows that

$B P = A C$, and from the similarity of triangles $P K B$ and $C K D - D C = \frac{1}{5} B P = \frac{1}{5} A C, \quad K C = \frac{1}{6} P C = \frac{1}{3} C E = \frac{5}{3}$.

From the right triangle $K D C$, we have $D C^{2} = K C^{2} - D K^{2} = \frac{16}{3}$. Therefore, $A D = 4 D C = \frac{16}{3}$. From the right triangle $A D B$,

$A B^{2} = A D^{2} + B D^{2} = \left(\frac{16}{3}\right)^{2} + 6^{2} = \frac{4}{9} \cdot 145$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-17.jpg?height=526&width=554&top_left_y=615&top_left_x=755)

## Answer

$\frac{2 \sqrt{145}}{3}$.