I give an analytic proof using the projective basis (see ProjectiveBase.html ) consisting of the points {A,B,C,D}, later being the coordinator or unit point (i.e. representing the point with coordinates (1,1,1)). Assume that line L is given in this system by an equation of the form

ax + by + cz = 0.

Then the following calculations are straightforward using essentially only the vector-product as indicated in each case:

1) line(B

2) A* = (1, -1, -1)x(a,b,c) => (b-c, -(a+c), a+b).

3) line(C

4) B* = (1,-1,1)x(a,b,c) => (-(b+c), a-c, a+b).

5) line(A

6) C* = (1,1-1)x(a,b,c) = (b+c, -(a+c), b-a).

7) line(A,A*) = (1, 0, 0)x(b-c, -(a+c), a+b) = (0, a+b, a+c) (the tripple standing for the coefficients of the line).

8) line(B,B*) = (0, 1, 0)x(-(b+c), a-c, a+b) = (a+b, 0, b+c).

9) line(C,C*) = (0, 0, 1)x(b+c, -(a+c), b-a) = (a+c, b+c, 0).

10) C

11) A

12) B

13) line(C,C

14) line(A,A

15) line(B,B

The determinant of the coefficients of the three last lines is zero since the matrix of coefficients is antisymmetric 3x3:

This implies that the three lines {AA

Let A'B'C' be the cevian triangle of point D with respect to triangle ABC. Draw from the vertices of ABC parallels to the sides of the triangle to form triangle A

The proof follows from the property of the previous section by applying it to the case in which the line L is the line at infinity, hence the points {A*, B*, C*} appearing there are here the "common" points of the pairs of parallels {(B'C',B

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