# Problem 7. (4 points)

A plane intersects the edges of the tetrahedron $ABCD$ emanating from vertex $B$ and divides them in the ratio $2:1$ (not necessarily from vertex $B$). This plane also intersects the lines $CD$ and $AC$ at points $E$ and $F$. Find the ratio of the areas of triangles $EFC$ and $ACD$.
Answer: $1: 9$ or $16: 9$.

Solution:

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

There are two possible cases. In the first case, $B C^{\prime}: C^{\prime} C=D^{\prime} D: D^{\prime} B=2: 1$. By Menelaus' theorem, we have $\frac{C E}{E D} \cdot \frac{D D^{\prime}}{D^{\prime} B} \cdot \frac{B C^{\prime}}{C^{\prime} C}=1$, from which $\frac{C E}{E D}=\frac{1}{4}$. In this case, point $E$ lies on the ray $D C$ beyond point $C$, so $\frac{C E}{C D}=\frac{1}{3}$. Similarly, $\frac{C F}{C A}=\frac{1}{3}$, and the ratio of the areas is $1: 9$ (since the angles at vertex $C$ of the triangles in question are equal).

In the second case, $B C^{\prime}: C^{\prime} C=D^{\prime} D: D^{\prime} B=1: 2$. By Menelaus' theorem, we have $\frac{C E}{E D} \cdot \frac{D D^{\prime}}{D^{\prime} B} \cdot \frac{B C^{\prime}}{C^{\prime} C}=1$, from which $\frac{C E}{E D}=\frac{4}{1}$. In this case, point $E$ lies on the ray $C D$ beyond point $D$, so $\frac{C E}{C D}=\frac{4}{3}$. Similarly, $\frac{C F}{C A}=\frac{4}{3}$, and the ratio of the areas is $16: 9$ (since the angles at vertex $C$ of the triangles in question are equal).