The cross ratio has an interpretation in terms of the angles of lines EA, EB, EC, ED at E. The dependence of E(A,B,C,D) from these angles shows that a second line (g) intersecting the four lines EA, EB, EC, ED defines the same cross ratio.

(I) If two lines e, e' carry each four points with the same cross ratio, (ABCD)=(A'B'C'D') say, and two corresponding points coincide (B=B' say i.e. e, e' intersect at B=B'). Then the lines joining corresponding points pass through the same point E.

(II) If two line pencils E(ABCD), E'(A'B'C'D') have a common line (f.e. EC = E'C') then the other pairs of homologous lines intersect at three collinear points.

The two propositions are dual to each other and trivial to prove. Consider f.e. in the second property line e of the intersection points of pairs (EA, E'A') and (EB, E'B'). Then by the invariance of the cross ratio on e, the intersection point of e with ED must coincide with that of E'D'. First property can serve as basis for a simple proof of Pascal's theorem on hexagons inscribed in conics (see Pascal.html ).

One prominent example of harmonic pencil consists of an angle and its two bisectors.

Another prominent example is the one related to a complete quadrilateral ABCD. This is created by fixing four points in general position. It has six sides resulting by all combinations of the points by two. It defines also three additional points (below labeled by {E,F,G}) named

A third example of harmonic pencil of four lines is the one formed by two sides {AB,AC} of a triangle ABC, the corresponding

In particular, if the lines are given in the form { y = a

In particular, if the lines make a harmonic pencil, then last ratio is -1. There results an interesting transformation each time one isgiven two fixed lines, say determined by their corresponding coefficients {a

This is an involutive Moebius transformation (see Involution.html ) whose determinant is equal to (a

Pascal.html

Harmonic.html

Harmonic_Bundle.html

Involution.html

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