a line named

a circle, to be found in Pascal.html .

again the special proof for the circle.

Denote by (AB)=0 the equation of the line joining points A and B. Since the conic passes through

points {A,B,C,D} it can be represented in the form

(AB)(CD)-(BC)(AD)=0.

For the same reason i.e. because it passes through {D,E,F,A} it can be represented in the form

(DE)(FA)-(EF)(AD)=0.

Equating these expressions we obtain

(AB)(CD)-(DE)(FA) = ((BC)-(EF))(AD).

Left hand side represents a conic circumcscribing quadrilateral (q) with sides {AB,DE,CD,AF} i.e.

passing through points {A,D,P,Q}. The right hand side is a product of lines one of which is the

diagonal (AD)=0 of (q). Hence the other factor ((BC)-(EF))=0 represents the other diagonal

(PQ)=0 of (q).

But from its form follows that it passes also through the intersection point of lines (BC)=0 and

(EF)=0, which is R.

to cubic curves and the following simple fact for them:

c

The proof of Pascal's theorem follows by applying this theorem to the two (degenerate) cubics c

consisting from lines c

The two cubics intersect at nine points: the 6 vertices of the hexagon and the 3 intersections of opposite

sides. Consider then the cubic c

three intersections of opposite sides, say points P and Q. Then, by the aforementioned theorem this cubic

will pass also from the ninth intersection point which is R necessarily lying on line PQ.

the 6th point on any line passing through one of these five points. Thus through this theorem,

starting by five arbitrary points we can construct arbitrary many points on the conic defined by

these five points.

Given {A,B,C,D,E} draw a (variable) line a through A and find its intersection I with CD. Let G

be the intersection of lines (AB, DE). Draw line GI and find its intersection K with line BC. The

sixth point F on line a must be on line KE according to Pascal.

consecutive vertices coincide, a case where the line joining two points, must be replaced with the

tangent at the (double) point. This is the case with the figure below, where at J we take the

tangent to the conic.

The figure gives also a recipe for the geometric construction of the tangent at a point J of the

conic, using points on the conic, lines and their intersections:

We define four other points I, H, G and L and join as shown. All dotted lines are known, and

points M, O are immediately constructed. Then N is determined as intersection point of MO and

GH. From this follows the construction of the tangent, by joining N and J.

See the references below for some other aspects of Pascal's theorem.

[ChaslesApercu] Chasles, M.

[SalmonConics] Salmon, G.

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