I am very new at analytic geometry and this question may be very easy but I am literally stuck.
Let $SABC$ be a tetrahedron, and let $A’,B’,C’$ be arbitrary points of the edges $SA,SB,SC$ respectively. Let $M$ be the intersection point of the planes $A’BC, B’CA, C’AB$, and $N$ be the intersection point of the planes $AB’C’, BC’A’,CA’B’$. Show that M, N and S are collinear points.
Is it enough to show that M and N are scaler times S? I am not sure about what should I do and I cannot imagine the figure.
Sorry for this question and thanks in advance for any help.
Define $D = C'N\cap SAB(A'B')$
Because of plane $C'NA'B$, $DA'B$ is a line, same for $DB'A$
Meaning $D=B'A\cap A'B$
Notice the same argument implies $CM\cap SAB(A'B')=B'A\cap A'B = C'N\cap SAB(A'B')$
meaning $CM$ intersects $C'N$ which implies $CC'MN$ is a plane which contains $S$ because of line $CC'S$. by the same argument we have plane $BB'MNS$. the intersection of these two planes is the $MNS$ line desired.