6. Given a rectangle $A B C D$. Point $M$ is the midpoint of side $A B$, point $K$ is the midpoint of side $B C$. Segments $A K$ and $C M$ intersect at point $E$. How many times smaller is the area of quadrilateral $M B K E$ compared to the area of quadrilateral $A E C D$?
Answer: 4 times.

Solution. Draw segments $MK$ and $AC$. Quadrilateral $MBKE$ consists of triangles $MBK$ and $MKE$, while quadrilateral $AEC D$ consists of triangles $AEC$ and $ACD$. We can reason in different ways.

1st method. Triangles $MBK$ and $ACD$ are right-angled, and the legs of the first are half the length of the legs of the second, so they are similar, and the area of triangle $ACD$ is 4 times the area of triangle $MBK$.

Since $M$ and $K$ are the midpoints of $AB$ and $BC$ respectively, $MK$ is the midline of triangle $ABC$, so $MK \parallel AC$ and $MK = 0.5 AC$. From the parallelism of lines $MK$ and $AC$, it follows that triangles $MKE$ and $AEC$ are similar, and since the similarity coefficient is 0.5, the area of triangle $AEC$ is 4 times the area of triangle $MKE$.

Now: $S_{AEC D} = S_{AEC} + S_{ACD} = 4 S_{MKE} + 4 S_{MBK} = 4 (S_{MKE} + S_{MBK}) = 4 S_{MBKE}$.

2nd method. Let the area of rectangle $ABCD$ be $S$. Then the area of triangle $ACD$ is $\frac{1}{2} S$ (the diagonal of the rectangle divides it into two equal triangles), and the area of triangle $MBK$ is $\frac{1}{2} MB \cdot BK = \frac{1}{2} \cdot \frac{1}{2} AB \cdot \frac{1}{2} BC = \frac{1}{8} AB \cdot BC = \frac{1}{8} S$.

Since $M$ and $K$ are the midpoints of segments $AB$ and $BC$, $AK$ and $CM$ are medians of triangle $ABC$, so $E$ is the centroid of triangle $ABC$, i.e., the distance from $E$ to $AC$ is $\frac{1}{3} h$, where $h$ is the height of triangle $ABC$ from vertex $B$. Then the area of triangle $AEC$ is $\frac{1}{2} AC \left(\frac{1}{3} h\right) = \frac{1}{3} \left(\frac{1}{2} AC h\right) = \frac{1}{3} S_{ABC} = \frac{1}{3} \left(\frac{1}{2} S\right) = \frac{1}{6} S$. Therefore, for the area of quadrilateral $AEC D$, which is the sum of the areas of triangles $AEC$ and $ACD$, we get: $\frac{1}{2} S + \frac{1}{6} S = \frac{2}{3} S$.

Furthermore, since $MK$ is the midline of triangle $ABC$, the area of triangle $MKE$ is $\frac{1}{2} MK \cdot \left(\frac{1}{2} h - \frac{1}{3} h\right) = \frac{1}{2} \left(\frac{1}{2} AC\right) \left(\frac{1}{6} h\right) = \frac{1}{12} \left(\frac{1}{2} AC h\right) = \frac{1}{12} S_{ACD} = \frac{1}{24} S$. Therefore, for the area of quadrilateral $MBKE$, which is the sum of the areas of triangles $MBK$ and $MKE$, we get: $\frac{1}{8} S + \frac{1}{24} S = \frac{1}{6} S$.

Thus, the ratio of the areas of quadrilaterals $AEC D$ and $MBKE$ is $\frac{2}{3} S : \left(\frac{1}{6} S\right) = 4$.

Grading criteria.

- Correct solution and correct answer - 7 points.
- Correct solution, but incorrect answer due to an arithmetic error - 5 points.