On 22-4-2012, at 7.48 AM, Luís Lopes wrote:
> Now I wonder how to prove (synthetically?) that
> (ABCD) = (A'B'C'D')
> Please see
Pappus originally proved it in his Lemmas to the Porisms of Euclid, beginning with Lemma III ("Collection" Book VII Prop.129). He considered several cases allowing for parallels to occur. (Modern use of the line at infinity reduces them to one case.) His proofs also introduced various parallels in order to use Euclid VI.2.
Later writers adapted the proof in various ways, includng the introduction of the sine function as in the web site which Luis mentioned. In an attempt to ease things for students, I developed my own proof from the Theorem of Menelaus. This may be in the literature somewhere, but I haven't seen it. If any parallels occur, ratios of the form PN/NQ with N at infinity are replaced by -1 as usual, which replaces Menelaus by Euclid VI.2. Hence cases with parallels are included in the proof below.
Pappus Lemma III is the special case where A = A', so the ranges (ABCD) and (AB'C'D') are perspective from O (the point called M in the web site diagram). Apply the Theorem of Menelaus twice to triangle ABB' using transversals ODD' and OCC'.
(BD/DA)(AD'/D'B')(B'O/OB) = -1 = (BC/CA)(AC'/C'B')(B'O/OB).
Cancelling the factor B'O/OB gives (BD/DA)(AD'/D'B') = (BC/CA)(AC'/C'B').
Some algebraic rearranging and removing four minus signs then gives
(AC.BD)/(AD.BC) = (AC'.B'D')/(AD'.B'C'), i.e. the cross-ratio (AB,CD) = (AB',C'D').
Next, permuting the points of both ranges shows that the cross-ratio of A, B, C, D in any order is equal to the cross-ratio of A, B', C' D' in the corresponding order. Hence the common point (such as A above) need not be named first.
Finally, the general result comes from moving crabwise. Suppose now that the ranges (ABCD) and (A'B'C'D') are perspective from O.
Let A'D meet OBB' at X and OCC' at Y.
Then the cross-ratio (AB,CD) = (A'X, YD) by the above result with D common
= (A'B',C'D') by the above result with A' common.