Prismatoid plane section with straightedge and compass I came up with a task I'm out of ideas how to do a solution. Perhaps I'm not paying attention to an obvious thing, but still. All I came up with is to build orthogonal projections but that's not a good position of a solid to make these clear.

In this image, there is a prismatoid with two bases: bottom one is a pentagon $ABCDE$, and the top one is a triangle $A_1B_1C_1$. Its lateral faces include two triangles ($AEA_1$ and $BCB_1$) and three trapezoids ($ABB_1A_1$, $DEA_1C_1$, and $CDC_1B_1$). Point $M$ belongs to a plane $A_1B_1C_1$, point $N$ belongs to a plane $ABA_1$, point $K$ belongs to a plane $AEB_1$. The task is to dissect a prismatoid $ABCDEA_1B_1C_1$ with a plane $MNK$ using straightedge and compass, i.e. to construct each line where a plane $MNK$ intersects a prismatoid $ABCDEA_1B_1C_1$.
tl,dr: $ABCDE\parallel A_1B_1C_1, M\in(A_1B_1C_1), N\in(ABA_1), K\in(AEB_1)$. Use SE&C to construct lines $\{ \alpha_1,\alpha_2,\dots,\alpha_n \} \ni (MNK)\ \cap ABCDEA_1B_1C_1$.
Any tips are highly appreciated.
 A: Let $G$ be the intersection between line $KN$ and plane $A_1B_1C_1$: line $MG$ is then the intersection between planes $KMN$ and $A_1B_1C_1$. Its two intersections $P$, $Q$ with triangle $A_1B_1C_1$ are two vertices of the section to be constructed. This doesn't work if $K$ lies on $ABCDE$: construct in that case the midpoint $K'$ of $KM$ and do the construction with $K'$ instead of $K$.
Let $H$ be the intersection between line $KM$ and plane $ABCDE$: line $NH$ is then the intersection between planes $KMN$ and $ABCDE$. Its two intersections $R$, $S$ with pentagon $ABCDE$ are two more vertices of the section to be constructed. This doesn't work if $K$ lies on $A_1B_1C_1$: construct in that case the midpoint $K'$ of $KN$ and do the construction with $K'$ instead of $K$.
To construct the remaining vertices, consider again line $KN$: it must intersect one of the lateral faces, for instance $BCB_1$, at some point $I$. On face $BCB_1$ also lies a point already found ($R$, for instance): the intersection of line $RI$ with triangle $BCB_1$ will then give another vertex. Do the same with line $KM$, if needed, to complete the construction.

EDIT.
See here for the construction of the intersection between a plane and a line.
