On the legs AX, AY of a fixed angle take the fixed points {B,C} on AX and variable points {B',C'} on AY, such that B'A/B'C'=k with a constant k. Then the intersection point D of lines {B'C,C'B} lies on a line parallel to AY.
A simple proof can be given using the projective base defined by points {A,C,Y,G}, G being the centroid of triangle ACY.
Assume then B = A+mC fixed, B'=A+pY, C'=A+qY. The ratios of the oriented segments created by these points are (see BarycentricCoordinates2.html )
B'A/B'Y = -p and C'A/C'Y = -q.
Further, expressing ratio B'A/B'C' in terms of these ratios we have:
k = B'A/B'C = (1+q)/(1-(q/p)), which solved for q gives q = p(k-1)/(k+p) (*).
Line BC' has coefficients given (ibid) by the cross product (1,m,0)x(1,0,q)=(mq,-q,-m) and line B'C has coefficients (0,1,0)x(1,0,p)=(p,0,-1). Thus D is given by (mq,-q,-m)x(p,0,-1)=(q,m(q-p),pq).
Taking into account (*) we see that the coordinates of D are a multiple of (1-k, m(1+p), p(1-k)).
Equating these coordinates with (rx,ry,rz) and eliminating r and p we find that the homogeneous coordinates of D satisfy the equation:
mx + (k-1)y + mz =0.
This is a line whose point at infinity is (m,k-1,m)x(1,1,1)=(k-1-m)(1,0,-1) i.e. the same with the point at infinity of the line y=0 which is AY. Thus the two lines are parallel.
Remark-1 Analogously one proves the fact that also points {E,F} of the figure move on lines parallel to AY as {B',C'} vary on this line so that B'A/B'C'=k (k constant). Remark-2 This innocent locus is a special case of a slightly more general configuration resulting by moving the origin of measurements on AY away from the intersection point A of lines {AX,AY}. i.e. considering a different from A point A' on AY and taking B'A'/B'C'=k. The analogous resulting locus is then a conic passing through points {A',B,C}.
![[alogo]](alogo.png){kind=small}
Line locus from ratio ![[0_0]](LineLocusRatio_files/LineLocusRatio_0_0_0.png)
![[0_1]](LineLocusRatio_files/LineLocusRatio_0_0_1.png)