The tetrahedron ABCD is given . The points M, N and K lie on the edges AD , AB and BC, respectively, and AM:MD = 2:3 , BN:AN = 1:2 and BK:KC = 1:1. Construct a section of the tetrahedron with the MNK plane. In what respect does this plane divide the edge CD?
P.S. It is forbidden to use Menelaus' theorem for a tetrahedron.
I connected the points N and K, M and N, K and P (the point on the edge of CD). Built a straight MP parallel to a straight NK. Thus, I built the MNK plane (MNKP). How to find the DP:PC I don't know.
Let the line through $M$ parallel to $NK$ intersect plane $BCD$ at $Q$ (see figure). The intersection $P$ between lines $QK$ and $CD$ is also the intersection of plane $KMN$ with $CD$.
Drop perpendicular lines $AH$, $ML$, $NI$ to plane $BCD$. We have $ML={3\over5}AH$ and $NI={1\over3}AH$, hence $$ ML={9\over5}NI \quad\text{and}\quad MQ={9\over5}NK. $$ Construct $F$ on line $BD$ such that $MF\parallel AB$. We have $$ MF={3\over5}AB={9\over5}NB. $$ Hence triangles $MQF$ and $NKB$ are similar, $FQ\parallel BC$ and $$ FQ={9\over5}BK={9\over10}BC. $$ If $FQ$ intersects $CD$ at $E$, then $FE={3\over5}BC$ and $$ QE=FQ-FE={3\over10}BC={3\over5}KC. $$ Finally, $PC={5\over3}PE={5\over8}EC$ and $EC={2\over5}DC$, that is: $$ PC={1\over4}DC. $$