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:
2. Multiplying the three
identities, we get:
3. Using the intersecting
secant theorem, we get:
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.
Blaise Pascal
(16231662) French mathematician, philosopher and
inventor, discovered his famous theorem at the age of
16, in 1640, and produced a treatise on conic sections
entitled Essai pour les coniques.
See also:
Geometric art of the Pascal's theorem
