According to Menelaus' Theorem ([12, Section 3.4]) there holds [CN/DN] x [DA/AP] x [BP/BC] = 1 (*), and according to Ceva's Theorem ([12, Section 1.2]) there holds [CK/KD] x [DA/AP] x [BP/BC] = 1 (**).
It is important to note that there is a preference and a higher chance of successful handling of different tasks if one is acquainted with many theorems, such as Menelaus' Theorem, Ceva's Theorem, Pascal's Theorem and others, which not each student and even teacher knows.
In the triangle [DELTA][AA.sub.1]B, cut by the transversal [CC.sub.1], we'll apply the Menelaus' theorem:
In the triangle [DELTA][BB.sub.1]C, cut by the transversal [AA.sub.1], we apply again the Menelaus' theorem:
We apply one more time the Menelaus' theorem in the triangle [DELTA][CC.sub.1]A cut by the transversal [BB.sub.1]:
We use Menelaus' theorem
for the sides of the triangle ABC cut by the line [A.sub.1][B.sub.1] (fig.
We described about generalizations of Menelaus' theorem to polygons and polyhedrons, and about backward generalization, from polyhedrons to polygons in chapter 3.
Also in chapter 3, since Menelaus' theorem is based on measurements of the line segments.
The idea of "cycle" takes central role in discovery of tetrahedral Menelaus' theorem. Though, Euler's contribution on graph theory tells us that we can't just loop through all edges for only once.
This equation is what we will now say Menelaus' theorem for tetrahedrons.