Pappus's hexagon theorem

<h2 id="proof-affine-form">Proof: affine form</h2>

If the affine form of the statement can be proven, then the projective form of Pappus's theorem is proven, as the extension of a pappian plane to a projective plane is unique. 
Because of the parallelity in an affine plane one has to distinct two cases: 
 
 
 
 g
 ∦
 h
 
 
 {\displaystyle g\not \parallel h}
 
 and 
 
 
 
 g
 ∥
 h
 
 
 {\displaystyle g\parallel h}
 
. The key for a simple proof is the possibility for introducing a "suitable" coordinate system:
Case 1: The lines 
 
 
 
 g
 ,
 h
 
 
 {\displaystyle g,h}
 
 intersect at point 
 
 
 
 S
 =
 g
 ∩
 h
 
 
 {\displaystyle S=g\cap h}
 
.
In this case coordinates are introduced, such that 
 
 
 
 
 S
 =
 (
 0
 ,
 0
 )
 ,
 
 A
 =
 (
 0
 ,
 1
 )
 ,
 
 c
 =
 (
 1
 ,
 0
 )
 
 
 
 {\displaystyle \;S=(0,0),\;A=(0,1),\;c=(1,0)\;}
 
 (see diagram).

B
 ,
 C
 
 
 {\displaystyle B,C}
 
 have the coordinates 
 
 
 
 
 B
 =
 (
 0
 ,
 γ
 )
 ,
 
 C
 =
 (
 0
 ,
 δ
 )
 ,
 
 γ
 ,
 δ
 ∉
 {
 0
 ,
 1
 }
 
 
 {\displaystyle \;B=(0,\gamma ),\;C=(0,\delta ),\;\gamma ,\delta \notin \{0,1\}}
 
.
From the parallelity of the lines 
 
 
 
 B
 c
 ,
 
 C
 b
 
 
 {\displaystyle Bc,\;Cb}
 
 one gets 
 
 
 
 b
 =
 (
 
 
 
 δ
 γ
 
 
 
 ,
 0
 )
 
 
 {\displaystyle b=({\tfrac {\delta }{\gamma }},0)}
 
 and the parallelity of the lines 
 
 
 
 A
 b
 ,
 B
 a
 
 
 {\displaystyle Ab,Ba}
 
 yields 
 
 
 
 a
 =
 (
 δ
 ,
 0
 )
 
 
 {\displaystyle a=(\delta ,0)}
 
. Hence line 
 
 
 
 C
 a
 
 
 {\displaystyle Ca}
 
 has slope 
 
 
 
 −
 1
 
 
 {\displaystyle -1}
 
 and is parallel line 
 
 
 
 A
 c
 
 
 {\displaystyle Ac}
 
.
Case 2: 
 
 
 
 g
 ∥
 h
  
 
 
 {\displaystyle g\parallel h\ }
 
 (little theorem).
In this case the coordinates are chosen such that 
 
 
 
 
 c
 =
 (
 0
 ,
 0
 )
 ,
 
 b
 =
 (
 1
 ,
 0
 )
 ,
 
 A
 =
 (
 0
 ,
 1
 )
 ,
 
 B
 =
 (
 γ
 ,
 1
 )
 ,
 
 γ
 ≠
 0
 
 
 {\displaystyle \;c=(0,0),\;b=(1,0),\;A=(0,1),\;B=(\gamma ,1),\;\gamma \neq 0}
 
. From the parallelity of 
 
 
 
 A
 b
 ∥
 B
 a
 
 
 {\displaystyle Ab\parallel Ba}
 
 and 
 
 
 
 c
 B
 ∥
 b
 C
 
 
 {\displaystyle cB\parallel bC}
 
 one gets 
 
 
 
 
 C
 =
 (
 γ
 +
 1
 ,
 1
 )
 
 
 
 {\displaystyle \;C=(\gamma +1,1)\;}
 
 and 
 
 
 
 
 a
 =
 (
 γ
 +
 1
 ,
 0
 )
 
 
 
 {\displaystyle \;a=(\gamma +1,0)\;}
 
, respectively, and at least the parallelity 
 
 
 
 
 A
 c
 ∥
 C
 a
 
 
 
 {\displaystyle \;Ac\parallel Ca\;}
 
.

<h2 id="proof-with-homogeneous-coordinates">Proof with homogeneous coordinates</h2>
Choose homogeneous coordinates with

C
 =
 (
 1
 ,
 0
 ,
 0
 )
 ,
 
 c
 =
 (
 0
 ,
 1
 ,
 0
 )
 ,
 
 X
 =
 (
 0
 ,
 0
 ,
 1
 )
 ,
 
 A
 =
 (
 1
 ,
 1
 ,
 1
 )
 
 
 {\displaystyle C=(1,0,0),\;c=(0,1,0),\;X=(0,0,1),\;A=(1,1,1)}
 
.
On the lines 
 
 
 
 A
 C
 ,
 A
 c
 ,
 A
 X
 
 
 {\displaystyle AC,Ac,AX}
 
, given by 
 
 
 
 
 x
 
 2
 
 
 =
 
 x
 
 3
 
 
 ,
 
 
 x
 
 1
 
 
 =
 
 x
 
 3
 
 
 ,
 
 
 x
 
 2
 
 
 =
 
 x
 
 1
 
 
 
 
 {\displaystyle x_{2}=x_{3},\;x_{1}=x_{3},\;x_{2}=x_{1}}
 
, take the points 
 
 
 
 B
 ,
 Y
 ,
 b
 
 
 {\displaystyle B,Y,b}
 
 to be

B
        =
        (
        p
        ,
        1
        ,
        1
        )
        ,
        
        Y
        =
        (
        1
        ,
        q
        ,
        1
        )
        ,
        
        b
        =
        (
        1
        ,
        1
        ,
        r
        )
      
    
    {\displaystyle B=(p,1,1),\;Y=(1,q,1),\;b=(1,1,r)}

for some 
 
 
 
 p
 ,
 q
 ,
 r
 
 
 {\displaystyle p,q,r}
 
. The three lines 
 
 
 
 X
 B
 ,
 C
 Y
 ,
 c
 b
 
 
 {\displaystyle XB,CY,cb}
 
 are 
 
 
 
 
 x
 
 1
 
 
 =
 
 x
 
 2
 
 
 p
 ,
 
 
 x
 
 2
 
 
 =
 
 x
 
 3
 
 
 q
 ,
 
 
 x
 
 3
 
 
 =
 
 x
 
 1
 
 
 r
 
 
 {\displaystyle x_{1}=x_{2}p,\;x_{2}=x_{3}q,\;x_{3}=x_{1}r}
 
, so they pass through the same point 
 
 
 
 a
 
 
 {\displaystyle a}
 
 if and only if 
 
 
 
 r
 q
 p
 =
 1
 
 
 {\displaystyle rqp=1}
 
. The condition for the three lines 
 
 
 
 C
 b
 ,
 c
 B
 
 
 {\displaystyle Cb,cB}
 
 and 
 
 
 
 X
 Y
 
 
 {\displaystyle XY}
 
 with equations 
 
 
 
 
 x
 
 2
 
 
 =
 
 x
 
 1
 
 
 q
 ,
 
 
 x
 
 1
 
 
 =
 
 x
 
 3
 
 
 p
 ,
 
 
 x
 
 3
 
 
 =
 
 x
 
 2
 
 
 r
 
 
 {\displaystyle x_{2}=x_{1}q,\;x_{1}=x_{3}p,\;x_{3}=x_{2}r}
 
 to pass through the same point 
 
 
 
 Z
 
 
 {\displaystyle Z}
 
 is 
 
 
 
 r
 p
 q
 =
 1
 
 
 {\displaystyle rpq=1}
 
. So this last set of three lines is concurrent if all the other eight sets are because multiplication is commutative, so 
 
 
 
 p
 q
 =
 q
 p
 
 
 {\displaystyle pq=qp}
 
. Equivalently, 
 
 
 
 X
 ,
 Y
 ,
 Z
 
 
 {\displaystyle X,Y,Z}
 
 are collinear.
The proof above also shows that for Pappus's theorem to hold for a projective space over a division ring it is both sufficient and necessary that the division ring is a (commutative) field. German mathematician <a href="/facts/Gerhard_Hessenberg/R1khycWT">Gerhard Hessenberg</a> proved that Pappus's theorem implies <a href="/facts/Desargues%27s_theorem/vJzx0uPh">Desargues's theorem</a>.<a class="footnote-ref" id="fnref:4" href="#fn:4">4</a><a class="footnote-ref" id="fnref:5" href="#fn:5">5</a> In general, Pappus's theorem holds for some projective plane if and only if it is a projective plane over a commutative field. The projective planes in which Pappus's theorem does not hold are <a href="/facts/Desarguesian/vJzx0uPh">Desarguesian</a> projective planes over noncommutative division rings, and <a href="/facts/Non-Desarguesian_plane/G7rlClxR">non-Desarguesian planes</a>.
The proof is invalid if 
 
 
 
 C
 ,
 c
 ,
 X
 
 
 {\displaystyle C,c,X}
 
 happen to be collinear. In that case an alternative proof can be provided, for example, using a different projective reference.

<h2 id="dual-theorem">Dual theorem</h2>
Because of the <a href="/facts/Duality_(projective_geometry)/dxrjhRbB">principle of duality for projective planes</a> the dual theorem of Pappus is true:
If 6 lines 
 
 
 
 A
 ,
 b
 ,
 C
 ,
 a
 ,
 B
 ,
 c
 
 
 {\displaystyle A,b,C,a,B,c}
 
 are chosen alternately from two <a href="/facts/Pencil_(mathematics)/b75bj3wd">pencils</a> with centers 
 
 
 
 G
 ,
 H
 
 
 {\displaystyle G,H}
 
, the lines

X
        :=
        (
        A
        ∩
        b
        )
        (
        a
        ∩
        B
        )
        ,
      
    
    {\displaystyle X:=(A\cap b)(a\cap B),}

Y
        :=
        (
        c
        ∩
        A
        )
        (
        C
        ∩
        a
        )
        ,
      
    
    {\displaystyle Y:=(c\cap A)(C\cap a),}

Z
        :=
        (
        b
        ∩
        C
        )
        (
        B
        ∩
        c
        )
      
    
    {\displaystyle Z:=(b\cap C)(B\cap c)}

are concurrent, that means: they have a point 
 
 
 
 U
 
 
 {\displaystyle U}
 
 in common.
The left diagram shows the projective version, the right one an affine version, where the points

G
 ,
 H
 
 
 {\displaystyle G,H}
 
 are points at infinity. If point 
 
 
 
 U
 
 
 {\displaystyle U}
 
 is on the line 
 
 
 
 G
 H
 
 
 {\displaystyle GH}
 
 than one gets the "dual little theorem" of Pappus' theorem.

If in the affine version of the dual "little theorem" point 
 
 
 
 U
 
 
 {\displaystyle U}
 
 is a point at infinity too, one gets <a href="/facts/Thomsen%27s_theorem/z2PWWKMx">Thomsen's theorem</a>, a statement on 6 points on the sides of a triangle (see diagram). The Thomsen figure plays an essential role coordinatising an axiomatic defined projective plane.<a class="footnote-ref" id="fnref:6" href="#fn:6">6</a> The proof of the closure of Thomsen's figure is covered by the proof for the "little theorem", given above. But there exists a simple direct proof, too:
Because the statement of Thomsen's theorem (the closure of the figure) uses only the terms connect, intersect and parallel, the statement is <a href="/facts/Affine_map/BD7yDr10">affinely</a> invariant, and one can introduce coordinates such that

P
 =
 (
 0
 ,
 0
 )
 ,
 
 Q
 =
 (
 1
 ,
 0
 )
 ,
 
 R
 =
 (
 0
 ,
 1
 )
 
 
 {\displaystyle P=(0,0),\;Q=(1,0),\;R=(0,1)}
 
 (see right diagram). The starting point of the sequence of chords is 
 
 
 
 (
 0
 ,
 λ
 )
 .
 
 
 {\displaystyle (0,\lambda ).}
 
 One easily verifies the coordinates of the points given in the diagram, which shows: the last point coincides with the first point.

<h2 id="other-statements-of-the-theorem">Other statements of the theorem</h2>

In addition to the above characterizations of Pappus's theorem and its dual, the following are equivalent statements:

<ul><li>If the six vertices of a hexagon lie alternately on two lines, then the three points of intersection of pairs of opposite sides are collinear.<a class="footnote-ref" id="fnref:7" href="#fn:7">7</a></li>
<li>Arranged in a matrix of nine points (as in the figure and description above) and thought of as evaluating a <a href="/facts/Permanent_(mathematics)/eJWgdahx">permanent</a>, if the first two rows and the six "diagonal" triads are collinear, then the third row is collinear.</li></ul>

|
          
            
              
                
                  A
                
                
                  B
                
                
                  C
                
              
              
                
                  a
                
                
                  b
                
                
                  c
                
              
              
                
                  X
                
                
                  Y
                
                
                  Z
                
              
            
          
          |
        
      
    
    {\displaystyle \left|{\begin{matrix}A&B&C\\a&b&c\\X&Y&Z\end{matrix}}\right|}

That is, if 
 
 
 
  
 A
 B
 C
 ,
 a
 b
 c
 ,
 A
 b
 Z
 ,
 B
 c
 X
 ,
 C
 a
 Y
 ,
 X
 b
 C
 ,
 Y
 c
 A
 ,
 Z
 a
 B
  
 
 
 {\displaystyle \ ABC,abc,AbZ,BcX,CaY,XbC,YcA,ZaB\ }
 
 are lines, then Pappus's theorem states that 
 
 
 
 X
 Y
 Z
 
 
 {\displaystyle XYZ}
 
 must be a line. Also, note that the same matrix formulation applies to the dual form of the theorem when 
 
 
 
 (
 A
 ,
 B
 ,
 C
 )
 
 
 {\displaystyle (A,B,C)}
 
 etc. are triples of concurrent lines.<a class="footnote-ref" id="fnref:8" href="#fn:8">8</a>
<ul><li>Given three distinct points on each of two distinct lines, pair each point on one of the lines with one from the other line, then the joins of points not paired will meet in (opposite) pairs at points along a line.<a class="footnote-ref" id="fnref:9" href="#fn:9">9</a></li>
<li>If two triangles are <a href="/facts/Perspective_(geometry)/IKUbVuYw">perspective</a> in at least two different ways, then they are perspective in three ways.<a class="footnote-ref" id="fnref:10" href="#fn:10">10</a></li>
<li>If 
 
 
 
 
 A
 B
 ,
 C
 D
 ,
 
 
 
 {\displaystyle \;AB,CD,\;}
 
 and 
 
 
 
 E
 F
 
 
 {\displaystyle EF}
 
 are concurrent and 
 
 
 
 D
 E
 ,
 F
 A
 ,
 
 
 {\displaystyle DE,FA,}
 
 and 
 
 
 
 B
 C
 
 
 {\displaystyle BC}
 
 are concurrent, then 
 
 
 
 A
 D
 ,
 B
 E
 ,
 
 
 {\displaystyle AD,BE,}
 
 and 
 
 
 
 C
 F
 
 
 {\displaystyle CF}
 
 are concurrent.<a class="footnote-ref" id="fnref:11" href="#fn:11">11</a></li></ul>
<h2 id="origins">Origins</h2>
In its earliest known form, Pappus's Theorem is Propositions 138, 139, 141, and 143 of Book VII of <a href="/facts/Pappus_of_Alexandria/UABLMu0m">Pappus's</a> Collection.<a class="footnote-ref" id="fnref:12" href="#fn:12">12</a> These are Lemmas XII, XIII, XV, and XVII in the part of Book VII consisting of lemmas to the first of the three books of <a href="/facts/Euclid/dHpVyL4R">Euclid</a>'s Porisms.
The lemmas are proved in terms of what today is known as the cross ratio of four collinear points. Three earlier lemmas are used. The first of these, Lemma III, has the diagram below (which uses Pappus's lettering, with G for Γ, D for Δ, J for Θ, and L for Λ).

Here three concurrent straight lines, AB, AG, and AD, are crossed by two lines, JB and JE, which concur at J. 
Also KL is drawn parallel to AZ.
Then

KJ : JL :: (KJ : AG & AG : JL) :: (JD : GD & BG : JB).
These proportions might be written today as equations:<a class="footnote-ref" id="fnref:13" href="#fn:13">13</a>

KJ/JL = (KJ/AG)(AG/JL) = (JD/GD)(BG/JB).
The last compound ratio (namely JD : GD & BG : JB) is what is known today as the <a href="/facts/Cross_ratio/tj4DK3Vk">cross ratio</a> of the collinear points J, G, D, and B in that order; it is denoted today by (J, G; D, B). So we have shown that this is independent of the choice of the particular straight line JD that crosses the three straight lines that concur at A. In particular

(J, G; D, B) = (J, Z; H, E).
It does not matter on which side of A the straight line JE falls. In particular, the situation may be as in the next diagram, which is the diagram for Lemma X.

Just as before, we have (J, G; D, B) = (J, Z; H, E). Pappus does not explicitly prove this; but Lemma X is a converse, namely that if these two cross ratios are the same, and the straight lines BE and DH cross at A, then the points G, A, and Z must be collinear.
What we showed originally can be written as (J, ∞; K, L) = (J, G; D, B), with ∞ taking the place of the (nonexistent) intersection of JK and AG. Pappus shows this, in effect, in Lemma XI, whose diagram, however, has different lettering:

What Pappus shows is DE.ZH : EZ.HD :: GB : BE, which we may write as

(D, Z; E, H) = (∞, B; E, G).
The diagram for Lemma XII is:

The diagram for Lemma XIII is the same, but BA and DG, extended, meet at N. In any case, considering straight lines through G as cut by the three straight lines through A, (and accepting that equations of cross ratios remain valid after permutation of the entries,) we have by Lemma III or XI

(G, J; E, H) = (G, D; ∞ Z).
Considering straight lines through D as cut by the three straight lines through B, we have

(L, D; E, K) = (G, D; ∞ Z).
Thus (E, H; J, G) = (E, K; D, L), so by Lemma X, the points H, M, and K are collinear. That is, the points of intersection of the pairs of opposite sides of the hexagon ADEGBZ are collinear.
Lemmas XV and XVII are that, if the point M is determined as the intersection of HK and BG, then the points A, M, and D are collinear. That is, the points of intersection of the pairs of opposite sides of the hexagon BEKHZG are collinear.

<h2 id="notes">Notes</h2>

<ul><li><a href="/facts/Harold_Scott_MacDonald_Coxeter/D2l5jWGG">Coxeter, Harold Scott MacDonald</a> (1969), Introduction to Geometry (2nd ed.), New York: <a href="/facts/John_Wiley_%26_Sons/53g3LqJc">John Wiley & Sons</a>, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-471-50458-0, <a href="/facts/MR_(identifier)/uP137L11">MR</a> <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=0123930">0123930</a></li>
<li>Cronheim, A. (1953), "A proof of Hessenberg's theorem", Proceedings of the American Mathematical Society, 4 (2): 219–221, <a href="/facts/Doi_(identifier)/muM9Etpq">doi</a>:<a href="https://doi.org/10.2307%2F2031794">10.2307/2031794</a>, <a href="/facts/JSTOR_(identifier)/YTeVmaJ7">JSTOR</a> <a href="https://www.jstor.org/stable/2031794">2031794</a></li>
<li>Dembowski, Peter (1968), Finite Geometries, Berlin: Springer-Verlag</li>
<li>Heath, Thomas (1981) [1921], A History of Greek Mathematics, New York: Dover Publications</li>
<li>Hessenberg, Gerhard (1905), "Beweis des Desarguesschen Satzes aus dem Pascalschen", Mathematische Annalen, 61 (2), Berlin / Heidelberg: Springer: 161–172, <a href="/facts/Doi_(identifier)/muM9Etpq">doi</a>:<a href="https://doi.org/10.1007%2FBF01457558">10.1007/BF01457558</a>, <a href="/facts/ISSN_(identifier)/DPAflDvU">ISSN</a> <a href="https://search.worldcat.org/issn/1432-1807">1432-1807</a>, <a href="/facts/S2CID_(identifier)/ldJsHa2Y">S2CID</a> <a href="https://api.semanticscholar.org/CorpusID:120456855">120456855</a></li>
<li>Hultsch, Fridericus (1877), Pappi Alexandrini Collectionis Quae Supersunt, Berlin{{citation}}: CS1 maint: location missing publisher (link)</li>
<li>Kline, Morris (1972), Mathematical Thought From Ancient to Modern Times, New York: Oxford University Press</li>
<li>Pambuccian, Victor; Schacht, Celia (2019), "The axiomatic destiny of the theorems of Pappus and Desargues", in Dani, S. G.; Papadopoulos, A. (eds.), Geometry in history, Springer, pp. 355–399, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-3-030-13611-6</li>
<li>Whicher, Olive (1971), Projective Geometry, Rudolph Steiner Press, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0-85440-245-4</li></ul>
<h2 id="external-links">External links</h2>
<ul><li><a href="http://www.cut-the-knot.org/pythagoras/Pappus.shtml">Pappus's hexagon theorem</a> at <a href="/facts/Cut-the-knot/WLgV43jd">cut-the-knot</a></li>
<li><a href="http://www.cut-the-knot.org/Curriculum/Geometry/PappusDual.shtml">Dual to Pappus's hexagon theorem</a> at <a href="/facts/Cut-the-knot/WLgV43jd">cut-the-knot</a></li>
<li><a href="https://www-m10.ma.tum.de/foswiki/pub/Lehre/WS0809/GeometrieKalkueleWS0809/ch1.pdf">Pappus’s Theorem: Nine proofs and three variations</a></li></ul>

<h2 id="references">References</h2>

<ol>
<li id="fn:1">Coxeter, pp. 236–7 <a href="#fnref:1" class="footnote-back-ref">↩</a></li>
<li id="fn:2">Rolf Lingenberg: Grundlagen der Geometrie, BI-Taschenbuch, 1969, p. 93 <a href="#fnref:2" class="footnote-back-ref">↩</a></li>
<li id="fn:3">However, this does occur when 
 
 
 
 A
 B
 C
 
 
 {\displaystyle ABC}
 
 and 
 
 
 
 a
 b
 c
 
 
 {\displaystyle abc}
 
 are in perspective, that is, 
 
 
 
 A
 a
 ,
 B
 b
 
 
 {\displaystyle Aa,Bb}
 
 and 
 
 
 
 C
 c
 
 
 {\displaystyle Cc}
 
 are concurrent. <a href="/wiki/Perspective_(geometry)" target="_blank">/wiki/Perspective_(geometry)</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></li>
<li id="fn:4">Coxeter 1969, p. 238 - Coxeter, Harold Scott MacDonald (1969), Introduction to Geometry (2nd ed.), New York: John Wiley & Sons, ISBN 978-0-471-50458-0, MR 0123930 <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=0123930" target="_blank">https://mathscinet.ams.org/mathscinet-getitem?mr=0123930</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></li>
<li id="fn:5">According to (Dembowski 1968, pg. 159, footnote 1), Hessenberg's original proof Hessenberg (1905) is not complete; he disregarded the possibility that some additional incidences could occur in the Desargues configuration. A complete proof is provided by Cronheim 1953. - Dembowski, Peter (1968), Finite Geometries, Berlin: Springer-Verlag <a href="#fnref:5" class="footnote-back-ref">↩</a></li>
<li id="fn:6">W. Blaschke: Projektive Geometrie, Springer-Verlag, 2013, ISBN 3034869320, S. 190 <a href="/wiki/ISBN_(identifier)" target="_blank">/wiki/ISBN_(identifier)</a> <a href="#fnref:6" class="footnote-back-ref">↩</a></li>
<li id="fn:7">Coxeter, p. 231 <a href="#fnref:7" class="footnote-back-ref">↩</a></li>
<li id="fn:8">Coxeter, p. 233 <a href="#fnref:8" class="footnote-back-ref">↩</a></li>
<li id="fn:9">Whicher, chapter 14 <a href="#fnref:9" class="footnote-back-ref">↩</a></li>
<li id="fn:10">Coxeter 1969, p. 238 - Coxeter, Harold Scott MacDonald (1969), Introduction to Geometry (2nd ed.), New York: John Wiley & Sons, ISBN 978-0-471-50458-0, MR 0123930 <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=0123930" target="_blank">https://mathscinet.ams.org/mathscinet-getitem?mr=0123930</a> <a href="#fnref:10" class="footnote-back-ref">↩</a></li>
<li id="fn:11">Coxeter, p. 233 <a href="#fnref:11" class="footnote-back-ref">↩</a></li>
<li id="fn:12">Heath (Vol. II, p. 421) cites these propositions. The latter two can be understood as converses of the former two. Kline (p. 128) cites only Proposition 139. The numbering of the propositions is as assigned by Hultsch. <a href="#fnref:12" class="footnote-back-ref">↩</a></li>
<li id="fn:13">A reason for using the notation above is that, for the ancient Greeks, a ratio is not a number or a geometrical object. We may think of ratio today as an equivalence class of pairs of geometrical objects. Also, equality for the Greeks is what we might today call congruence. In particular, distinct line segments may be equal. Ratios are not equal in this sense; but they may be the same. <a href="#fnref:13" class="footnote-back-ref">↩</a></li>
</ol>

Pappus's hexagon theorem open-in-new

Pappus's hexagon theorem