§ 27: Projective pencils on a conic

If X, Y, A, B, C, D are points on a conic J, then J is the conic generated by the projectivity φ from X onto Y with φ(XA) = YA, φ(XB) = YB, φ(XC) = YC. We also have φ(XD) = YD.
Hence (XA, XB; XC, XD) = (YA, YB; YC, YD), so we can define:

Definition: The cross ratio of four points A, B, C, D on a conic J is the number (XA, XB; XC, XD), with X on J arbitrary.

Definition: A projectivity from a conic J onto itself is a bijection from J onto J that preserves cross ratio.

Fundamental Theorem: Given three points A, B, C on a conic J and three points A ', B ', C ' on J.
Then there exists exactly one projectivity φ: JJ with φ(A) = A ', φ(B) = B ', φ(C) = C '.

i) Existence:
Choose D on J. Let ψ be the projectivity from the pencil of lines D onto D with ψ(DA) = DA ', ψ(DB) = DB ', ψ(DC) = DC '. Define φ as follows: for X on J, let φ(X) be the intersection point unequal to D of ψ(DX) and J. Check that this φ meets the requirements.
ii) Unicity:
Suppose φ1 and φ2 both meet the requirements.
Then we have for X on J: (A, B; C, X) = (φ1(A), φ1(B); φ1(C), φ1(X)) = (A ', B ' ; C ', φ1(X)), and likewise: (A, B; C, X) = (A ', B ' ; C ', φ2(X)).
So (DA, DB; DC, DX) = (DA ', DB ' ; DC ', Dφ1(X)) = (DA ', DB ' ; DC ', Dφ2(X)).
So Dφ1(X)) = Dφ2(X)), and since φ1(X)) and φ2(X)) both lie on J, it follows that φ1(X) = φ2(X).

Note: If P is a point, then the mapping that, for X on J, maps X to the other intersection point of PX and J, is in general not a projectivity.

Theorem of Steiner: Let J be a conic and φ a projectivity from J onto J.
Then all the points Xφ(Y). Yφ(X), with X, Y on J, lie on one and the same line (the Pascal line of φ).

Proof: Choose three points A, B, C on J. We are going to prove that, for X and Y on J, Xφ(Y). Yφ(X) lies on the Pascal line of hexagon ABCφ(A)φ(B)φ(C).
Let ψ be the projectivity from the pencil of lines A onto the pencil of lines φ(A) with ψ(Aφ(B)) = φ(A)B, ψ(Aφ(C)) = φ(A)C, ψ(Aφ(A)) = φ(A)A.
Since Aφ(A) is invariant, this is a perspectivity whose axis is the line through Aφ(B)).φ(A)B and Aφ(C)).φ(A)C, this is the Pascal line of hexagon ABCφ(A)φ(B)φ(C).
Since (B, C; A, X) = (φ(B), φ(C); φ(A), φ(X)), we have (φ(A)B, φ(A)C; φ(A)A, φ(A)X)) = (Aφ(B), Aφ(C); Aφ(A), Aφ(X)).
So, since ψ preserves cross ratio, and because of O38, we find ψ(Aφ(X)) = φ(A)X.
This implies that Aφ(X). Xφ(A) lies on the Pascal line of hexagon ABCφ(A)φ(B)φ(C).
Likewise we deduce that Aφ(Y). Yφ(A) lies on that line.
Now if we apply the theorem of Pascal to hexagon AXYφ(A)φ(X)φ(Y), then we see that Xφ(Y). Yφ(X) also lies on the same line.

O102 Study the two texts in in the answers of this problem. Together, they form an application of the theory above.

O103 Given two triangles ABC and UWV. Construct triangle PQR whose sides go through U, V and W and whose vertices lie on the sides of triangle ABC.