Let A, B, C, D, E, F be points on a
circle and makes a hexagon out of them in an arbitrary order.
Then the three points L, M, N at which pairs of opposite sides
meet, lie on a straight line. This line is called the Pascal
line of the hexagon ABCDEF.
Proof:
1. Applying
Menelaus’ Theorem
respectively to transversal BCM, AFN, and DEL of triangle XYZ:

4. By substituting (3) into
(2), we get:
Thus, by the converse of
Menelaus' Theorem, points M, N, and L lie on a
transversal of triangle XYZ, therefore must be collinear.
Q.E.D.
