Theorem: Pappus's Hexagon Theorem

Let $A, B, C$ be a set of collinear points.
Let $a, b, c$ be another set of collinear points.
Let $X, Y, Z$ be the points of intersection of each of the straight lines $Ab$ and $aB$, $Ac$ and $aC$, and $Bc$ and $bC$.
Then $X, Y, Z$ are collinear points.

Proof:
300px
The notation has been changed to match the source.
Let $ACE$ be collinear and $BFD$ also be collinear.
Join $ABCDEF$ in order.
The points where opposite sides of the hexagon cut each other are $NLM$:
* $AB$ and $DE$ cross at $L$
* $BC$ and $EF$ cross at $N$
* $CD$ and $FA$ cross at $M$
$NLM$ are to be proved collinear.
There are two cases.
'''Case 1''':  $EF$ not $\parallel CD$.
Produce $EF$ to meet $CD$ at $U$.
Label points $V$ and $W$ in $\triangle UVW$.
Apply Menelaus's Theorem to five transversals of $\triangle UVW$.
{{begin-eqn}}
{{eqn | n = 1
      | l = \dfrac{UE}{EV} \cdot \dfrac {VA}{AW} \cdot \dfrac {WC}{CU}
      | r = -1
      | c = Menelaus's Theorem for $ECA$ in $\triangle ABC$
}}
{{eqn | n = 2
      | l = \dfrac{WC}{CU} \cdot \dfrac {UN}{NV} \cdot \dfrac {VB}{BW} 
      | r = -1
      | c = Menelaus's Theorem for $CNB$ in $\triangle ABC$
}}
{{eqn | n = 3
      | l = \dfrac{UE}{EV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WD}{DU} 
      | r = -1
      | c = Menelaus's Theorem for $ELD$ in $\triangle ABC$
}}
{{eqn | n = 4
      | l = \dfrac{UF}{FV} \cdot \dfrac {VA}{AW} \cdot \dfrac {WM}{MU} 
      | r = -1
      | c = Menelaus's Theorem for $AMF$ in $\triangle ABC$
}}
{{eqn | n = 5
      | l = \dfrac{UF}{FV} \cdot \dfrac {VB}{BW} \cdot \dfrac {WD}{DU} 
      | r = -1
      | c = Menelaus's Theorem for $BFD$ in $\triangle ABC$
}}
{{end-eqn}}
Multiply $(2) \cdot (3) \cdot (4)$:
{{begin-eqn}}
{{eqn | l = \dfrac{WC}{CU} \cdot \dfrac {UN}{NV} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VA}{AW} \cdot \dfrac {WM}{MU} 
      | r = -1
      | c = 
}}
{{eqn | n = 6
      | l = \dfrac {UN}{NV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WM}{MU} \cdot \paren {\dfrac{WC}{CU} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VA}{AW} }
      | r = -1
      | c = rearranging
}}
{{end-eqn}}
Multiply $(1) \cdot (5)$:
{{begin-eqn}}
{{eqn | l = \dfrac{UE}{EV} \cdot \dfrac {VA}{AW} \cdot \dfrac {WC}{CU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VB}{BW} \cdot \dfrac {WD}{DU}
      | r = 1
      | c =
}}
{{eqn | n = 7
      | l = \dfrac {WC}{CU} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV}  \cdot \dfrac {VA}{AW}
      | r = 1
      | c = rearranging
}}
{{end-eqn}}
Multiply $(7)$ by $-1$ and equate the {{LHS}} of $(6)$ to the {{LHS}} of $(7)$:
{{begin-eqn}}
{{eqn | l = \dfrac {UN}{NV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WM}{MU} \cdot \paren {\dfrac{WC}{CU} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VA}{AW} }
      | r = - \dfrac {WC}{CU} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV}  \cdot \dfrac {VA}{AW}
      | c =
}}
{{eqn | l = \dfrac {UN}{NV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WM}{MU} 
      | r = - 1
      | c =
}}
{{end-eqn}}
By Menelaus's Theorem, $NLM$ are collinear.
{{qed|lemma}}
'''Case 2''':  $EF \parallel CD$ (and $BC \parallel AF$).
300px
$EF \parallel CD$ gives similar triangles.
For the first two, $\angle B$ is shared.
{{begin-eqn}}
{{eqn | l = \angle BVN
      | r = \angle BWC
      | c = Parallelism implies Equal Corresponding Angles
}}
{{eqn | l = \triangle BVN
      | o = \sim
      | r = \triangle BWC
      | c = Triangles with Two Equal Angles are Similar
}}
{{eqn | ll = \leadsto
      | n = 1
      | l = \dfrac{VN}{BV} 
      | r = \dfrac{WC}{BW}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \angle BVF
      | r = \angle BWD
      | c = Parallelism implies Equal Corresponding Angles
}}
{{eqn | l = \triangle BVF
      | o = \sim
      | r = \triangle BWD
      | c = Triangles with Two Equal Angles are Similar
}}
{{eqn | ll = \leadsto
      | n = 2
      | l = \dfrac{BV}{VF} 
      | r = \dfrac{BW}{WD}
}}
{{end-eqn}}
For the next two, $\angle A$ is shared.
{{begin-eqn}}
{{eqn | l = \angle AWM
      | r = \angle AVF
      | c = Parallelism implies Equal Corresponding Angles
}}
{{eqn | l = \triangle AWM
      | o = \sim
      | r = \triangle AVF
      | c = Triangles with Two Equal Angles are Similar
}}
{{eqn | ll = \leadsto
      | n = 3
      | l = \dfrac{AW}{WM} 
      | r = \dfrac{AV}{VF}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \angle AWC
      | r = \angle AVE
      | c = Parallelism implies Equal Corresponding Angles
}}
{{eqn | l = \triangle AWC
      | o = \sim
      | r = \triangle AVE
      | c = Triangles with Two Equal Angles are Similar
}}
{{eqn | n = 4
      | l = \dfrac{WC}{AW} 
      | r = \dfrac{VE}{AV}
      | c
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \angle EVL
      | r = \angle DWL
      | c = Parallelism implies Equal Alternate Angles
}}
{{eqn | l = \angle ELV
      | r = \angle DLW
      | c = Vertical Angle Theorem
}}
{{eqn | l = \triangle EVL
      | o = \sim
      | r = \triangle DWL
      | c = Triangles with Two Equal Angles are Similar
}}
{{eqn | ll = \leadsto
      | n = 5
      | l = \dfrac{EV}{VL} 
      | r = \dfrac{DW}{WL}
}}
{{end-eqn}}
Multiply the {{LHS}} of $(1)$ through $(5)$ together as the new {{LHS}}.
Multiply the {{RHS}} of $(1)$ through $(5)$ together as the new {{RHS}}.
{{begin-eqn}}
{{eqn | n = 6
      | l = \dfrac {VN}{BV} \cdot \dfrac {BV}{VF} \cdot \dfrac {AW}{WM} \cdot \dfrac {WC}{AW} \cdot \dfrac {EV}{VL}  
      | r = \dfrac {WC}{BW} \cdot \dfrac {BW}{WD} \cdot \dfrac {AV}{VF} \cdot \dfrac {VE}{AV} \cdot \dfrac {DW}{WL}
      | c =
}}
{{eqn | l = \dfrac {VN}{VF} \cdot \dfrac {WC}{WM} \cdot \dfrac {EV}{VL}  
      | r = \dfrac {WC}{BW} \cdot \dfrac {BW}{WD} \cdot \dfrac {AV}{VF} \cdot \dfrac {VE}{AV} \cdot \dfrac {DW}{WL}
      | c = cancel $BV$ and $AW$
}}
{{eqn | l = \dfrac {VN}{VF} \cdot \dfrac {WC}{WM} \cdot \dfrac {EV}{VL}  
      | r = \dfrac {WC}{WD} \cdot \dfrac {VE}{VF} \cdot \dfrac {DW}{WL}
      | c = cancel $BW$ and $AV$
}}
{{eqn | l = \dfrac {VN}{VF} \cdot \dfrac {WC}{WM} \cdot \dfrac {VE}{VL}  
      | r = \dfrac {WC}{WD} \cdot \dfrac {VE}{VF} \cdot \dfrac {WD}{WL}
      | c = reverse $EV$ and $DW$
}}
{{eqn | l = \dfrac {VN}{VF} \cdot \dfrac {WC}{WM} \cdot \dfrac {VE}{VL}  
      | r = \dfrac {WC}{WL} \cdot \dfrac {VE}{VF}
      | c = cancel $WD$
}}
{{eqn | l = \dfrac {VN}{VL} \cdot \dfrac {WC}{WM}  
      | r = \dfrac {WC}{WL}
      | c = cancel $\dfrac {VE} {VF}$
}}
{{eqn | l = \dfrac {VN}{WM}
      | r = \dfrac {VL}{WL}
      | c = cancel $WC$ and rearrange
}}
{{eqn | l = \angle LVN
      | r = \angle LWM
      | c = Parallelism implies Equal Alternate Angles
}}
{{eqn | l = \triangle LVN
      | o = \sim
      | r = \triangle LWM
      | c = Triangles with One Equal Angle and Two Sides Proportional are Similar
}}
{{eqn | ll = \leadsto
      | l = \angle NLV
      | r = \angle MLW
}}
{{end-eqn}}
We have {{hypothesis}}:
:$VLW$ are collinear
Suppose $NLM$ are not collinear.
Then:
:$\angle NLV \ne \angle MLW$
But this is a contradiction, since $\triangle NLV \sim \triangle MLW$ and $\angle NLV = \angle MLW$.
Hence, $NLM$ are collinear.
The result follows.
{{qed}}