8,9

Let $M$ be the point of intersection of the medians of the base $ABC$ of the triangular prism $ABC A_1 B_1 C_1$; $N$ and $K$ be the points of intersection of the diagonals of the faces $AA_1C_1C$ and $BB_1C_1C$ respectively. The plane $MNK$ intersects the lines $B_1C_1$ and $CC_1$ at points $P$ and $Q$ respectively. Construct the section of the prism by the plane $MNK$ and find the ratios $B_1P : B_1C_1$ and $C_1Q : CC_1$.
The line $N K$ is parallel to the plane $A B C$, since the segment $N K$ is the midline of the triangle $A C 1 B$. Therefore, the intersecting plane intersects the plane $A B C$ along a line parallel to $N K$ and passing through the point $M$. Let $E$ and $F$ be the points of intersection of this line with $A C$ and $B C$ respectively, and the lines $E N$ and $F K$ intersect the edges $A 1 C 1$ and $B 1 C 1$ at points $G$ and $P$. Then the trapezoid $E F P G$ is the desired section. Let $H$ be the midpoint of $A B$. Since $E F \| A B$

$$
\frac{C H}{F^{\prime} B}=\frac{C M}{M H}=2
$$

Let $F B=a$. Then $C F=2 a$, and since $K$ is the point of intersection of the diagonals of the parallelogram $B C C 1 B 1, B 1 P=$ $C F=2 a$ and $C 1 P=F B=a$. Therefore,

$$
\frac{B_{1} P^{\prime}}{B_{1} C_{1}}=\frac{2 a}{3 a}=\frac{2}{3}
$$

Since $P C 1 \| F C$ and $F C=2 P C 1, P C 1$ is the midline of the triangle $F Q C$, therefore, $C 1 Q=C C 1$.

## Answer

$2: 3 ; 1: 1$