8,9 |  |

The height $B K$ of the rhombus $A B C D$, dropped to the side $A D$, intersects the diagonal $A C$ at point $M$. Find $M D$, if it is known that $B K=4, A K: K D=1: 2$.
![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-20.jpg?height=49&width=715&top_left_y=2874&top_left_x=0)

## Solution

Since $AC$ is the perpendicular bisector of diagonal $BD$, then $MD = MB$. Since $AM$ is the angle bisector of triangle $BAK$, then

$$
\frac{BM}{MK} = \frac{AB}{AK} = \frac{AD}{AK} = 3
$$

Therefore,

$$
MD = BM = \frac{3}{4} BK = 3
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-21.jpg?height=366&width=594&top_left_y=503&top_left_x=732)

Answer

3.