Answer: $1: 576$ or $625: 576$.

## Solution:

Let the points of intersection of the plane (which we will call $\alpha$) with the edges $D A, D B$, and $D C$ be $A^{\prime}, B^{\prime}$, and $C^{\prime}$, respectively. If $D A^{\prime}: A^{\prime} A = D B^{\prime}: B^{\prime} B$, the lines $A B$ and $A^{\prime} B^{\prime}$ are parallel, which means that the plane $\alpha$ does not intersect the line $A B$. Similar reasoning applies to the line $A C$. Therefore, $D A^{\prime}: A^{\prime} A \neq D B^{\prime}: B^{\prime} B = D C^{\prime}: C^{\prime} C$.

There are two possible cases. In the first case, $D A^{\prime}: A^{\prime} A = B^{\prime} B: D B^{\prime} = 5: 1$. By Menelaus' theorem, we have $\frac{A E}{E B} \cdot \frac{B B^{\prime}}{D B^{\prime}} \cdot \frac{D A^{\prime}}{A^{\prime} A} = 1$, from which $\frac{A E}{E B} = \frac{1}{25}$. In this case, point $E$ lies on the ray $B A$ beyond point $A$, so $\frac{A E}{A B} = \frac{1}{24}$. Similarly, $\frac{A F}{A C} = \frac{1}{24}$, and the ratio of the areas is $1: 576$ (since the sought triangles have equal angles at vertex $A$).

In the second case, $D A^{\prime}: A^{\prime} A = B^{\prime} B: D B^{\prime} = 1: 5$. By Menelaus' theorem, we have $\frac{A E}{E B} \cdot \frac{B B^{\prime}}{D B^{\prime}} \cdot \frac{D A^{\prime}}{A^{\prime} A} = 1$, from which $\frac{A E}{E B} = 25$. In this case, point $E$ lies on the ray $A B$ beyond point $B$, so $\frac{A E}{A B} = \frac{25}{24}$. Similarly, $\frac{A F}{A C} = \frac{25}{24}$, and the ratio of the areas is $625: 576$.
# Problem 7. (4 points)

A plane intersects the edges of the tetrahedron $ABCD$ emanating from vertex $D$ and divides them in the ratio $5:1$ (not necessarily from vertex $D$). This plane also intersects the lines $AB$ and $AC$ at points $E$ and $F$. Find the ratio of the areas of triangles $AEF$ and $ABC$.