Let L1, Ly, L2 be three planar lines that intersect each other at the same point P. Let Lx be a line perpendicular to Ly that intersects L1, Ly, and L2 at points A, O, and C respectively. Two rays emanating from O at an angle ‘g’ with respect to Lx, and symmetrically to Ly, intersect L1 and L2 at points B and D respectively, see Fig. 1.
1. Then lines L(C,B) and L(A,D) intersect each other at a point E that lies on Ly.
Moreover let B1 and D1 be the intersection points of L1 and L2 respectively with two other rays from O at an angle ‘g1’ with respect to Lx.
2. Then lines L(D1,B) and L( B1,D) also intersect each other at a point F that lies on Ly.
Fig.1
Proof. For the sake of simplicity let Lx and Ly be respectively the x- and y-axis of a rectangular coordinate system, and point P has coordinates [0,-1]. Let ‘a’ be the angle between L1 and Ly, and ‘b’ the angle between Ly and L2. We have
A = [tan(a), 0] ,
B = 1/(1-tan(a)*tan(g))*[tan(a), tan(a)*tan(g)] ,
B1= 1/(1-tan(a)*tan(g1))*[tan(a), tan(a)*tan(g1)] ,
C = [-tan(b), 0] ,
D = 1/(1-tan(b)*tan(g))*[-tan(b), tan(b)*tan(g)] ,
D1= 1/(1-tan(b)*tan(g1))*[-tan(b), tan(b)*tan(g1)] .
The line L(C,B) intersects Ly at a point E with the ordinate
Ey = (By-Cy)/(Bx-Cx)*(-Cx)+Cy , or
Ey = tan(a)*tan(b)*tan(g)/(tan(a)+tan(b)-tan(a)*tan(b)*tan(g)) .
The same expression for Ey we obtain for the line L(A,D). This proves the first statement.
The line L(D1,B) intersects Ly at a point F with the ordinate
Fy = (By-D1y)/(Bx-D1x)*(-D1x)+D1y , or
Fy = tan(a)*tan(b)*(tan(g)+tan(g1))/(tan(a)+tan(b)-tan(a)*tan(b)*(tan(g)+tan(g1))) .
The same expression for Fy we obtain for the line L(B1,D). This proves the second statement.
Let Q be the intersection point of line L(D,B) and Lx, then the abscise of Q is
Qx = -Dy*(Bx-Dx)/(By-Dy)+Dx , or
Qx = 2*tan(a)*tan(b)/(tan(b)-tan(a)) .
We observe that Qx is independent of ray angle ‘g’. This means that Q is a perspectivity point and points (P,A,B1,B) and (P,C,D1,D) have the same cross ratio. This cross ratio is
CR = PB1*AB/(PB*AB1) , or
CR = tan(g)/tan(g1) .
We observe that CR does not depend on the angles ‘a’ and ‘b’.