In a previous post, we were introduced to
Pascal's Hexagrammum Mysticum Theorem - a magical theorem - which states that if we draw a
hexagon inscribed in a
conic section then the three pairs of
opposite sides of the hexagon intersect at three points which lie on a
straight line.
For example, as in the following figure we have a hexagon inscribed in a
circle and the intersection points of the three pairs of the opposite sides of the hexagon $\{12, 45\}$, $\{23, 56\}$, $\{34, 61\}$ are collinear.
There is a useful tool to prove the
collinearity of points - the
Menelaus' Theorem - which states as follows:
Menelaus' Theorem: Given a triangle $ABC$ and three points $A'$, $B'$, $C'$ lying on the three lines $BC$, $CA$, $AB$, respectively. Then the three points $A'$, $B'$, $C'$ are collinear if and only if $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = 1.$$
Today, we will use
Menelaus' theorem to prove
Pascal's theorem for the circle case.