Projective Proof of Pascal's Theorem

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Projective Proof of Pascal's Theorem

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The applet illustrates a derivation of Pascal's theorem from Chasles' theorem.

I'll use the symbol "^" to denote the intersection of two lines, i.e. the point incident to the two lines. Similarly, the same symbol is used to denoted the line passing through two points, i.e., incident to the two points, with no ambiguity. Thus let A, B, C, D, E, F be six points on a conic c (ellipse in the applet above.) Introduce

P = AB ^ DE,
Q = CD ^ FA,
p = P ^ Q,
L, M = c ^ p,
R = EF ^ p,
S = BC ^ p.

T = AD ^ p is an auxiliary point useful in the proof. The idea is to show that R = S. (The proof that has been suggested by Hubert Shutrick is based on Chasles' theorem and is, therefore, entirely projective.) Using the cross-ratio notations for concurrent lines and collinear points, we have successively

(LQRM)	= F(LAEM)
	= D(LAEM)
	= (LTPM)
	= A(LDBM)
	= C(LDBM)
	= (LQSM)

And, since the value of the cross ratio and three collinear points define the fourth point uniquely, we see that indeed R = S.

Remark

Chasles' theorem has been used twice in the proof: passing from pencil F to pencil D and from pencil A to pencil C. However, if the points L, A, E, M in the first case and L, D, B, M in the second, are collinear, the identities hold directly from the definition of the cross-ratio thus obviating the need for Chasles' theorem. Thus the above derivation immediately applies to Pappus' theorem.

Pascal and Brianchon Theorems

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