8. In triangle $A B C$, sides $A B, A C, B C$ are equal to 5, 6, and 7 respectively. On the median $A M$, segment $A K$ equal to 2 is laid off. Find the ratio of the areas of triangles $A B K$ and $A C K$. Write the answer as $\frac{S_{A B K}}{S_{A C K}}$.
# Solution:

The median of a triangle divides the triangle into two equal-area (equal in area) triangles. In triangle $ABC$, the areas of triangles $ABM$ and $ACM$ are equal because $AM$ is its median. In triangle $KBC$, the segment $KM$ is the median, so the areas of triangles $KBM$ and $KCM$ are equal. $S_{ABK} = S_{ABM} - S_{\text{KBM}} = S_{ACM} - S_{\text{KCM}} = S_{ACK}$, thus we get $S_{ABK} : S_{ACK} = 1$.

Answer: 1.