8. In an acute-angled triangle $ABC$, a point $Q$ is chosen on side $AC$ such that $AQ: QC = 1: 2$.

From point $Q$, perpendiculars $QM$ and $QK$ are dropped to sides $AB$ and $BC$ respectively. It is given that $BM: MA = 4: 1, BK = KC$. Find $MK: AC$.
Solution. Draw the heights $A K_{1}$ and $C M_{1}$. The idea of the solution is as follows: we will show that triangles $M_{1} B K_{1}$, $M B K$, and $A B C$ are similar to each other; from this, it will be easy to find the required ratio.

Let's denote the lengths:

![](https://cdn.mathpix.com/cropped/2024_05_06_8f18e627c90a5374668dg-4.jpg?height=466&width=647&top_left_y=2026&top_left_x=1250)

$A Q=x, Q C=2 x, C K=z, K B=z, B M=4 y, M A=y$.

From the similarity of triangles $A K_{1} C$ and $Q K C$, we find

$$
K K_{1}=K C \cdot A Q / Q C=z / 2
$$

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Similarly, since $\triangle A Q M \sim \Delta A C M_{1}$, then

$$
M M_{1}=Q C \cdot A M / A Q=2 y
$$

Thus, since $B K_{1}=K K_{1}$ and $M M_{1}=M_{1} B$, then $\Delta M_{1} B K_{1} \sim \Delta M B K$ with a similarity coefficient of 2 (their common angle lies between proportional sides). It is well known that a triangle formed by two bases of heights and a vertex is similar to the original, specifically: $\triangle M_{1} B K_{1} \sim \triangle A B C$ with a similarity coefficient of $\cos \angle B$. Therefore, $M_{1} K_{1}: A C=\cos \angle B$ and then:

$$
M K: A C=2 \cos \angle B
$$

It remains to calculate $\cos \angle B$. The areas of similar triangles $\triangle M_{1} B K_{1}$ and $\triangle A B C$ are related as the square of the similarity coefficient:

$$
\frac{B M_{1} \cdot B K_{1}}{B A \cdot B C}=\frac{2 y \cdot \frac{z}{2}}{5 y \cdot 2 z}=\cos ^{2} \angle B
$$

From this, $\cos \angle B=\frac{1}{\sqrt{10}}$. Substituting the found value into (1), we get the answer.

Answer: $M K: A C=\frac{2}{\sqrt{10}}$.