# Pappus's Hexagon Theorem

## 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 \ \$ $\ds \angle NLV$ $=$ $\ds \angle MLW$
$VLW$ 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.

$\blacksquare$

## Also known as

Pappus's Hexagon Theorem is also known just as Pappus's Theorem, but there is more than one such.

Hence on $\mathsf{Pr} \infty \mathsf{fWiki}$ we rigorously use the full version.

## Source of Name

This entry was named for Pappus of Alexandria.

## Historical Note

Pappus's Hexagon Theorem was first proved by Pappus of Alexandria in about $300$ CE.

The theorem is stated as Propositions $138$, $139$, $141$, and $143$ of Book $\text{VII}$ of Pappus's Collection.

It is noted that it is a limiting case of Pascal's Mystic Hexagram.

In $1899$ its full significance was revealed by David Hilbert, during his work on clarifying the foundations of geometry.