
Pappus' Theorem
Today we will learn about Pappus' theorem. This theorem states that if we take three points 1, 3, 5 on a line, and another three points 2, 4, 6 on another line, then the three intersection points of the following line pairs \{12, 45\}, ~\{23, 56\}, ~\{34, 61\} are collinear.

Pascal's Theorem
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:
Today, we will use Menelaus' theorem to prove Pascal's theorem for the circle case.
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.
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