At the base of the pyramid $S A B C D$ lies a trapezoid $A B C D$ with bases $B C$ and $A D$, and $B C=2 A D$. Points $K$ and $L$ are taken on the edges $S A$ and $S B$, respectively, such that $2 S K=K A$ and $3 S L=L B$. In what ratio does the plane $K L C$ divide the edge $S D$?
## First Method.

The planes of the faces $SBC$ and $SAD$ pass through the parallel lines $BC$ and $AD$ and have a common point $S$. Therefore, they intersect along a line $l$, parallel to the bases of the trapezoid and passing through point $S$ (Fig.1). Let the line $CL$ intersect the line $l$ at point $P$, the line $PK$ intersect the line $SD$ at point $M$, and the line $AD$ at point $Q$. Denote $AD=a$. Then $BC=2a$. From the similarity of triangles $SLP$ and $BLC$, we find that

$$
SP=\frac{SL}{LB} \cdot BC=\frac{1}{3} \cdot 2a=\frac{2}{3}a
$$

and from the similarity of triangles $SKP$ and $AKQ$,

$$
AQ=\frac{AK}{KS} \cdot SP=2 \cdot \frac{2}{3}a=\frac{4}{3}a
$$

Then $QD=AQ-AD=\frac{4}{3}a-a=\frac{1}{3}a$, and from the similarity of triangles $SMP$ and $DMQ$ it follows that

$$
\frac{SM}{MD}=\frac{BP}{QD}=\frac{b^2}{\frac{2}{3}a}=2.
$$

## Second Method.

Let $M$ be the point of intersection of the cutting plane with the line $SD$ (Fig.2). Denote $V_{SABCD}=V, \frac{SM}{SD}=x$. Then

$$
\begin{gathered}
V_{SABC}=\frac{2}{3}V, V_{SADC}=\frac{1}{3}V, V_{SABD}=\frac{1}{3}V, V_{SBCD}=\frac{2}{3}V, \\
V_{SCKL}=\frac{SL}{SB} \cdot \frac{SK}{SA} \cdot \frac{SC}{SC} \cdot V_{SABC}=\frac{1}{4} \cdot \frac{1}{3} \cdot \frac{1}{1} \cdot \frac{2}{3}V, \\
V_{SCKM}=\frac{SK}{SA} \cdot \frac{SC}{SC} \cdot \frac{SM}{SD} \cdot V_{SACD}=\frac{1}{3} \cdot \frac{1}{1} \cdot x \cdot \frac{1}{3}V, \\
V_{SKLM}=\frac{SL}{SB} \cdot \frac{SK}{SA} \cdot \frac{SM}{SD} \cdot V_{SABD}=\frac{1}{4} \cdot \frac{1}{3} \cdot x \cdot \frac{1}{3}V, \\
V_{SCLM}=\frac{SL}{SB} \cdot \frac{SC}{SC} \cdot \frac{SM}{SD} \cdot V_{SBCD}=\frac{1}{4} \cdot \frac{1}{1} \cdot x \cdot \frac{2}{3}V.
\end{gathered}
$$

The condition for the points $K, L, C$, and $M$ to lie in the same plane is equivalent to the equality

$$
V_{SCKL}+V_{SCKM}=V_{SKLM}+V_{SCLM},
$$

From this, we find that $\frac{SM}{SD}=x=\frac{2}{3}$. Therefore, $SM:MD=2:1$.

## Answer

$2:1$