Pappus's Hexagon Theorem/Proof 1
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
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 $\triangle UVW$.
| \(\text {(1)}: \quad\) | \(\ds \dfrac{UE}{EV} \cdot \dfrac {VA}{AW} \cdot \dfrac {WC}{CU}\) | \(=\) | \(\ds -1\) | Menelaus's Theorem for $ECA$ in $\triangle ABC$ | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds \dfrac{WC}{CU} \cdot \dfrac {UN}{NV} \cdot \dfrac {VB}{BW}\) | \(=\) | \(\ds -1\) | Menelaus's Theorem for $CNB$ in $\triangle ABC$ | ||||||||||
| \(\text {(3)}: \quad\) | \(\ds \dfrac{UE}{EV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WD}{DU}\) | \(=\) | \(\ds -1\) | Menelaus's Theorem for $ELD$ in $\triangle ABC$ | ||||||||||
| \(\text {(4)}: \quad\) | \(\ds \dfrac{UF}{FV} \cdot \dfrac {VA}{AW} \cdot \dfrac {WM}{MU}\) | \(=\) | \(\ds -1\) | Menelaus's Theorem for $AMF$ in $\triangle ABC$ | ||||||||||
| \(\text {(5)}: \quad\) | \(\ds \dfrac{UF}{FV} \cdot \dfrac {VB}{BW} \cdot \dfrac {WD}{DU}\) | \(=\) | \(\ds -1\) | Menelaus's Theorem for $BFD$ in $\triangle ABC$ |
Multiply $(2) \cdot (3) \cdot (4)$:
| \(\ds \dfrac{WC}{CU} \cdot \dfrac {UN}{NV} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VA}{AW} \cdot \dfrac {WM}{MU}\) | \(=\) | \(\ds -1\) | ||||||||||||
| \(\text {(6)}: \quad\) | \(\ds \dfrac {UN}{NV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WM}{MU} \cdot \paren {\dfrac{WC}{CU} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VA}{AW} }\) | \(=\) | \(\ds -1\) | rearranging |
Multiply $(1) \cdot (5)$:
| \(\ds \dfrac{UE}{EV} \cdot \dfrac {VA}{AW} \cdot \dfrac {WC}{CU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VB}{BW} \cdot \dfrac {WD}{DU}\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\text {(7)}: \quad\) | \(\ds \dfrac {WC}{CU} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VA}{AW}\) | \(=\) | \(\ds 1\) | rearranging |
Multiply $(7)$ by $-1$ and equate the left hand side of $(6)$ to the left hand side of $(7)$:
| \(\ds \dfrac {UN}{NV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WM}{MU} \cdot \paren {\dfrac{WC}{CU} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VA}{AW} }\) | \(=\) | \(\ds - \dfrac {WC}{CU} \cdot \dfrac {VB}{BW} \cdot \dfrac{UE}{EV} \cdot \dfrac {WD}{DU} \cdot \dfrac{UF}{FV} \cdot \dfrac {VA}{AW}\) | ||||||||||||
| \(\ds \dfrac {UN}{NV} \cdot \dfrac {VL}{LW} \cdot \dfrac {WM}{MU}\) | \(=\) | \(\ds - 1\) |
By Menelaus's Theorem, $NLM$ are collinear.
$\Box$
Case 2: $EF \parallel CD$ (and $BC \parallel AF$).
$EF \parallel CD$ gives similar triangles.
For the first two, $\angle B$ is shared.
| \(\ds \angle BVN\) | \(=\) | \(\ds \angle BWC\) | Parallelism implies Equal Corresponding Angles | |||||||||||
| \(\ds \triangle BVN\) | \(\sim\) | \(\ds \triangle BWC\) | Triangles with Two Equal Angles are Similar | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \dfrac{VN}{BV}\) | \(=\) | \(\ds \dfrac{WC}{BW}\) |
| \(\ds \angle BVF\) | \(=\) | \(\ds \angle BWD\) | Parallelism implies Equal Corresponding Angles | |||||||||||
| \(\ds \triangle BVF\) | \(\sim\) | \(\ds \triangle BWD\) | Triangles with Two Equal Angles are Similar | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \dfrac{BV}{VF}\) | \(=\) | \(\ds \dfrac{BW}{WD}\) |
For the next two, $\angle A$ is shared.
| \(\ds \angle AWM\) | \(=\) | \(\ds \angle AVF\) | Parallelism implies Equal Corresponding Angles | |||||||||||
| \(\ds \triangle AWM\) | \(\sim\) | \(\ds \triangle AVF\) | Triangles with Two Equal Angles are Similar | |||||||||||
| \(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \dfrac{AW}{WM}\) | \(=\) | \(\ds \dfrac{AV}{VF}\) |
| \(\ds \angle AWC\) | \(=\) | \(\ds \angle AVE\) | Parallelism implies Equal Corresponding Angles | |||||||||||
| \(\ds \triangle AWC\) | \(\sim\) | \(\ds \triangle AVE\) | Triangles with Two Equal Angles are Similar | |||||||||||
| \(\text {(4)}: \quad\) | \(\ds \dfrac{WC}{AW}\) | \(=\) | \(\ds \dfrac{VE}{AV}\) |
| \(\ds \angle EVL\) | \(=\) | \(\ds \angle DWL\) | Parallelism implies Equal Alternate Angles | |||||||||||
| \(\ds \angle ELV\) | \(=\) | \(\ds \angle DLW\) | Vertical Angle Theorem | |||||||||||
| \(\ds \triangle EVL\) | \(\sim\) | \(\ds \triangle DWL\) | Triangles with Two Equal Angles are Similar | |||||||||||
| \(\text {(5)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \dfrac{EV}{VL}\) | \(=\) | \(\ds \dfrac{DW}{WL}\) |
Multiply the left hand side of $(1)$ through $(5)$ together as the new left hand side.
Multiply the right hand side of $(1)$ through $(5)$ together as the new right hand side.
| \(\text {(6)}: \quad\) | \(\ds \dfrac {VN}{BV} \cdot \dfrac {BV}{VF} \cdot \dfrac {AW}{WM} \cdot \dfrac {WC}{AW} \cdot \dfrac {EV}{VL}\) | \(=\) | \(\ds \dfrac {WC}{BW} \cdot \dfrac {BW}{WD} \cdot \dfrac {AV}{VF} \cdot \dfrac {VE}{AV} \cdot \dfrac {DW}{WL}\) | |||||||||||
| \(\ds \dfrac {VN}{VF} \cdot \dfrac {WC}{WM} \cdot \dfrac {EV}{VL}\) | \(=\) | \(\ds \dfrac {WC}{BW} \cdot \dfrac {BW}{WD} \cdot \dfrac {AV}{VF} \cdot \dfrac {VE}{AV} \cdot \dfrac {DW}{WL}\) | cancel $BV$ and $AW$ | |||||||||||
| \(\ds \dfrac {VN}{VF} \cdot \dfrac {WC}{WM} \cdot \dfrac {EV}{VL}\) | \(=\) | \(\ds \dfrac {WC}{WD} \cdot \dfrac {VE}{VF} \cdot \dfrac {DW}{WL}\) | cancel $BW$ and $AV$ | |||||||||||
| \(\ds \dfrac {VN}{VF} \cdot \dfrac {WC}{WM} \cdot \dfrac {VE}{VL}\) | \(=\) | \(\ds \dfrac {WC}{WD} \cdot \dfrac {VE}{VF} \cdot \dfrac {WD}{WL}\) | reverse $EV$ and $DW$ | |||||||||||
| \(\ds \dfrac {VN}{VF} \cdot \dfrac {WC}{WM} \cdot \dfrac {VE}{VL}\) | \(=\) | \(\ds \dfrac {WC}{WL} \cdot \dfrac {VE}{VF}\) | cancel $WD$ | |||||||||||
| \(\ds \dfrac {VN}{VL} \cdot \dfrac {WC}{WM}\) | \(=\) | \(\ds \dfrac {WC}{WL}\) | cancel $\dfrac {VE} {VF}$ | |||||||||||
| \(\ds \dfrac {VN}{WM}\) | \(=\) | \(\ds \dfrac {VL}{WL}\) | cancel $WC$ and rearrange | |||||||||||
| \(\ds \angle LVN\) | \(=\) | \(\ds \angle LWM\) | Parallelism implies Equal Alternate Angles | |||||||||||
| \(\ds \triangle LVN\) | \(\sim\) | \(\ds \triangle LWM\) | Triangles with One Equal Angle and Two Sides Proportional are Similar | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \angle NLV\) | \(=\) | \(\ds \angle MLW\) |
By by 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.
$\blacksquare$
Sources
- 1967: H.S.M. Coxeter and S.L. Greitzer: Geometry Revisited: Chapter $3$ : "Collinearity and Concurrence."