Suppose that points $A,B$ and $C$ lie on the same straight line, and
$D,E$ and $F$ lie on *another* one. Draw line segments
$AE,BD,BF,CE,AF$ and $CD$.

Let $P$ be the intersection point of $AE$
and $BD$.

Let $Q$ be the intersection point of $AF$
and $CD$.

Let $R$ be the intersection point of $BF$
and $CE$.

Then $P,Q$ and $R$
are *collinear*. (i.e. The red dashed line
and the red solid line are collinear.)

## Discussion

To prove that three points are collinear, Menelaus’s Theorem is often
useful. To apply the theorem to this problem, (TL;DR;C image below)
one needs to find a triangle such that *exactly* one of points $P,Q$ and $R$ lies on
the *line* formed by two vertices of the triangle, but *not* on the
*segment* connecting those two vertices, while each of the remaining
two points lies on a segment whose extension *doesn’t* contain any one
of points $P,Q$ and $R$.

It is obvious that points $Q$ and $R$ divide some segments internally. However,
$P$ *also* has such property. How can one
find a segment through $P$ such that $P$ *externally* divides the segment?

The answer can be find out by a careful observation of the two
segments that pass through $P$. Since we
want a segment so that $P$ *externally*
divides the segment, we need at least two points on a line through
$P$. Moreover, those two points should be
*on the same side of the line relative to $P$*. This led us to define a *new* point. Let
$S$ be the intersection of $AF$ and $BD$.

Then the light blue triangle $\triangle B\color{blue}{S}F$ is what we need.

## The proof

Since I have been tired of drawing figures using TikZ and typing
equations with $\rm \LaTeX$, I *don’t* want to illustrate or prove any
one of the four cases of Co-side Theorem, on which this proof heavily
depends. I refer readers to Zhang J. Z.’s work.^{1}

From the above figure, it’s clear that what we need to show is

**The main idea of this proof is simple: reduce everything to what is
given.** Thus, points $P,Q,R$ and $S$ (i.e. red and blue points) should be eventually *eliminated*.
The tool that we’re going to use is Co-side Theorem.

On RHS of (1), one can see that the red points are gone, and we still need to remove
the $S$. Moreover, dealing with length of
segments is *easier* than their area because the former and the later
are one-dimensional and two-dimensional quantities respectively.
Therefore, we rearrange the area of the triangles to get

However, those two triangles in (2) have *no* common side.
Thus, we need to work out the area of the two coloured triangles in
the figure below. (i.e. $\triangle CD\color{blue}{S}$ and $\triangle
AE\color{blue}{S}$)

From equation (2), observe that $AB/CB$ and $EF/DF$ are independent from each other. Then, we need to relate the area of those two colored triangles to the ratios of segment lengths in (2).

I still use Co-side Theorem since I *can’t* think of another theorem
to eliminate S. In order to use that
theorem, one has to find another triangle and compare it with one of
those two coloured triangles. I’ve looked at the picture and the
symbols to find out some clues. Since the letter $C$ is present in
both $AB/CB$ and $\triangle CD\color{blue}{S}$, $C$ *can’t* be the
“common side” of $\triangle CD\color{blue}{S}$ and the triangle that
we want. Looking at $\triangle CD\color{blue}{S}$ again, it’s obvious
that $D\color{blue}{S}$ is the “common side” because $D$,
$S$ and $B$ are collinear.

The $CB/BA$ in (3) cancels the $AB/CB$ in (2). Therefore, we’re almost done!

Substituting (3) and (4) into (2), we get

Hence, by Menelaus’s Theorem, we can conclude that
$P,Q$ and $R$ are
*collinear*, and prove Pappus’s Hexagon Theorem