PP_{1} ----- = (P_{1},P_{2},P) = the dividing ratio of point P relative to (P_{1},P_{2}) PP_{2}Remark: If (P_{1},P_{2},P) < 0, point P is between P_{1} and P_{2}.
In a cartesian coordinate system we take P(x,y); P_{1}(x_{1},y_{1}); P_{2}(x_{2},y_{2}). Then we know :
x_{1} - k x_{2} x = ------------ and 1 - k y_{1} - k y_{2} y = ------------- 1 - kSolving these equations for k, we find
x - x_{1} k = -------- and x - x_{2} y - y_{1} k = --------- y - y_{2}We also know that the ideal point of P_{1}P_{2} has a dividing ratio = 1.
cross ratio of the ordered points A,B,C,D (A,B,C) = --------- (A,B,D)We denote the cross ratio of the ordered points A,B,C,D as (A,B,C,D).
(A,B,C) CA DA x_{3} - x_{1} x_{4} - x_{1} (A,B,C,D) = -------- = ---- : ---- = --------- : ---------- (A,B,D) CB DB x_{3} - x_{2} x_{4} - x_{2} y_{3} - y_{1} y_{4} - y_{1} = -------- : ---------- y_{3} - y_{2} y_{4} - y_{2}
(A,B,C,D) = -1 <=> x_{3} - x_{1} x_{4} - x_{1} -------- : -------- = -1 x_{3} - x_{2} x_{4} - x_{2} <=> x_{3} - x_{1} x_{4} - x_{1} -------- = - ------- x_{3} - x_{2} x_{4} - x_{2} <=> (x_{3} - x_{1})(x_{4} - x_{2}) = - (x_{4} - x_{1})(x_{3} - x_{2}) <=> ... <=> 2(x_{1} x_{2} + x_{3} x_{4}) = (x_{1} + x_{2})(x_{3} + x_{4})Similarly, we find
(A,B,C,D) = -1 <=> 2(y_{1} y_{2} + y_{3} y_{4}) = (y_{1} + y_{2})(y_{3} + y_{4})
(A,B,C,D) = -1 <=> (A,B,C) -------- = -1 (A,B,D) <=> (A,B,C) = - (A,B,D) <=> (A,B,C) or (A,B,D) is < 0 <=> C or D is between A and B (exclusive or)
(A,B,C,D) = -1 <=> (B,A,C,D) = -1 <=> (A,B,D,C) = -1 <=> (B,A,D,C) = -1 (A,B,C,D) = -1 <=> (C,D,A,B) = -1 <=> (D,C,B,A) = -1
then x_{2} = - x_{1} and y_{2} = - y_{1} and so (A,B,C,D) = -1 <=> 2(x_{1} x_{2} + x_{3} x_{4}) = (x_{1} + x_{2})(x_{3} + x_{4}) <=> 2(-x_{1} x_{1} + x_{3} x_{4}) = 0 <=> x_{1}^{2} = x_{3} x_{4} Similarly y_{1}^{2} = y_{3} y_{4}
(A,B,C,D) = -1 <=> 2(x_{1} x_{2} + x_{3} x_{4}) = (x_{1} + x_{2})(x_{3} + x_{4}) <=> 2(x_{3} x_{4}) = x_{2}(x_{3} + x_{4}) <=> 2 1 1 --- = --- + ---- x_{2} x_{3} x_{4} <=> x_{2} is the harmonic mean of x_{3} and x_{4}
C(x_{1} + h x_{2}, y_{1} + h y_{2}, z_{1} + h z_{2}) D(x_{1} + h' x_{2}, y_{1} + h' y_{2}, z_{1} + h' z_{2}) Then: (A,B,C) = x_{1} + h x_{2} x_{1} --------- - ---- z_{1} + h z_{2} z_{1} h z_{2} ---------------- = ... = ... = ------ x_{1} + h x_{2} x_{2} z_{1} --------- - --- z_{1} + h z_{2} z_{2} Similarly - h' z_{2} (A,B,D) = -------- z_{1} And from these results h (A,B,C,D) = --- h'
A(x_{1},y_{1},z_{1}), B(x_{2},y_{2},z_{2}), C(x_{1} + h x_{2}, y_{1} + h y_{2}, z_{1} + h z_{2}), D(x_{1} + h' x_{2}, y_{1} + h' y_{2}, z_{1} + h' z_{2})Now BY DEFINITION :
h (A,B,C,D) = --- h'and all previous properties hold.
(A,B,C,D) = -1 <=> h' = - h <=> h + h' = 0
C(x_{1} + h x_{2}, y_{1} + h y_{2}, 1 + h) D(x_{1} + h' x_{2}, y_{1} + h' y_{2}, 1 + h') If h = -1 and h' = 1 then (A,B,C,D) = -1 but then we have C(x_{1} - x_{2}, y_{1} - y_{2}, 0) and D(x_{1} + x_{2}, y_{1} + y_{2}, 2) <=> x_{1} + x_{2} y_{1} + y_{2} C(x_{1} - x_{2}, y_{1} - y_{2}, 0) and D( --------,---------, 1) 2 2 <=> C is the ideal point of AB and D is the midpoint of [AB]Conclusion:
a: u_{1} x + v_{1} y + w_{1} z = 0 b: u_{2} x + v_{2} y + w_{2} z = 0 c: (u_{1} + h u_{2})x + (v_{1} + h v_{2})y + (w_{1} + h w_{2})z = 0 d: (u_{1} + h'u_{2})x + (v_{1} + h'v_{2})y + (w_{1} + h'w_{2})z = 0A line e intersects these lines respectively in A,B,C and D.
e: u_{o} x + v_{o} y + w_{o} z = 0 then A has coordinates (x_{1},y_{1},z_{1}) = | v_{o} w_{o} | | u_{o} w_{o} | | u_{o} v_{o} | ( | v_{1} w_{1} | , - | u_{1} w_{1} | , | u_{1} v_{1} | ) B has coordinates (x_{2},y_{2},z_{2}) = | v_{o} w_{o} | | u_{o} w_{o} | | u_{o} v_{o} | ( | v_{2} w_{2} | , - | u_{2} w_{2} | , | u_{2} v_{2} | ) C has coordinates | v_{o} w_{o} | | u_{o} w_{o} | | u_{o} v_{o} | ( | v_{1}+hv_{2} w_{1}+hw_{2}|, - | u_{1}+hu_{2} w_{1}+hw_{2}| , | u_{1}+hu_{2} v_{1}+hv_{2} | ) | v_{o} w_{o} | | v_{o} w_{o} | | u_{o} w_{o} | | u_{o} w_{o} | =( | v_{1} w_{1} | + h| v_{2} w_{2} | , - | u_{1} w_{1} |- h| u_{2} w_{2} | , | u_{o} v_{o} | | u_{o} v_{o} | | u_{1} v_{1} |+ h | u_{2} v_{2} | ) = (x_{1} + h x_{2}, y_{1} + h y_{2}, s1 + h z_{2}) and here the same h appears as in the line c! Similarly we find for D (x_{1} + h'x_{2}, y_{1} + h'y_{2}, s1 + h'z_{2}) and here the same h' appears as in the line dFrom this, we deduce an important corollary : Take an ordered quartet of concurrent lines (a,b,c,d) in the projective plane. The cross ratio (A,B,C,D) = k is independent of the choice of the line e! So, we can define
(a,b,c,d) = -1 <=> b and c are the bisectors of a and bThe proof is left as an exercise.
We'll prove that (a,b,c,d) = -1 and (A,B,C,D) = -1
(A,B,C,D) = (a,b,c,d) intersection with line A'D' = (A',B',C',D') = (S'A',S'B',S'C',S'D') = intersection with line AD = (B,A,C,D) Thus, k = (A,B,C,D) = (B,A,C,D) CA DA CB DB Now (A,B,C,D) = --- : ---- and (B,A,C,D) = --- : ---- CB DB CA DA So, 1 k = --- <=> k^{2} = 1 <=> k = -1 or k = 1 kBut a cross ratio = 1 is impossible for a quartet of points.
-1 = (a,b,c,d) = (A,B,C,D) = (A',B',C',D') = (S'A',S'B',S'C',S'D') -1 = (D'C,D'C',D'S',D'S) = ...