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COURSE OF PROJECTIVE GEOMETRY

§ 15:

*O59* Let *A*, *B*, *C*, *D* be the freely situated fixed points.So the intersection point *S*_{1} of *AB* and *CD* is a fixed point.

Then both of the lines *AB* and *CD* contain three fixed points, so they are pointwise invariant (FS §10). Likewise, each of the lines of the complete quadrangle *ABCD*
is pointwise invariant.

Now let *X* be an arbitrary point in *P*^{2}. Choose a line through *X* that intersects the complete quadrangle *ABCD* in at least three points.
Then this line contains three fixed points, so it is pointwise invariant. So *X* is a fixed pont. Since this holds for each *X*, the projective mapping is identity.

*O60* That φ induces a bijection θ from the pencil of lines *L* to the pencil of lines φ(*L*) follows from the definition of a collineation. Because φ
is a projective transformation, θ preserves cross ratio. According to the second proposition of §10, θ is a projectivity.

If φ leaves the two pencils linewise invariant,we easily find four freely situated fixed points,because the intersection point of two invariant lines is a fixed point. According to *O59*,
φ is identity.

*O61*
Let __c__ = __a__ + λ(__b__ - __a__), with λ outside {0,1}. Then *P*(__c__) = *P*(__a__) +
λ(*P*(__b__) - *P*(__a__)).

So *P*(__a__), *P*(__b__), *P*(__c__) are collinear and *P*(*A*)*P*(*B*) / *P*(*A*)*P*(*C*) =
||*P*(__a__) - *P*(__b__)|| / ||*P*(__a__) - *P*(__c__)|| = 1/|λ| =
||__a__ - __b__|| / ||__a__ - __c__|| = *AB* / *AC*.

Now if *A*, *B*, *C*, *D* are four points on a line *l* in {x_{3} = 1}, we find:

(*A*, *B*; *C*, *D*) = (*P(A)*, *P(B)*; *P(C)*, *P(D)*) = (φ(*A*), φ(*B*); φ(*C*), φ(*D*)) (projection of *P(l)* onto
φ(*l*) from *O*).

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