QA-Tf8: QA-Orthopolar Line

 
Let P1,P2,P3,P4 be a quadrilateral and let L be a random line.
Let X1,X2,X3,X4 be the Orthopoles (for definition see [13]) of L resp. wrt triangles (P2,P3,P4), (P3,P4,P1), (P4,P1,P2), (P1,P2,P3).
X1, X2, X3, X4 are collinear on a line being called the QA-Orthopolar Line, which is QA-Tf8(L).
 
QA Tf8 QA OrthopolarLine 01 
 
CT-Coordinates:
Let L=(l:m:n), then 1st coordinate of QA-Tf8(L) is:
- q r (l SB - n SB + l SC - m SC) (SA SB + SA SC + SB SC) l (m - n)
+p r (m SA - n SA - l SC + m SC) (m2 SA2 - 2 m n SA2 + n2 SA2 + l2 SA SB + l m SA SB - 2 l n SA SB - m n SA SB + n2 SA SB + l2 SA SC - l m SA SC + m2 SA SC - m n SA SC + l m SB SC - m n SB SC)
+p q (m SA - n SA + l SB - n SB) (m2 SA2 - 2 m n SA2 + n2 SA2 + l2 SA SB - l n SA SB - m n SA SB + n2 SA SB + l2 SA SC - 2 l m SA SC + m2 SA SC + l n SA SC - m n SA SC + l n SB SC - m n SB SC)
 
Properties:
QA-Tf6(L) lies on QA-Tf8(L). See Ref-34, QFG#1186.
• The Orthopole X of the Anticomplementary Triangle of QA-Tr1 lies on QA-Tf8(L). See Ref-34, Ngo Quang Duong, QFG#1229. When L= QL-L5 (Eulerline of QA-Tr1), then X = a point on QA-Ci1, the circumcircle of QA-Tr1. Actually X = ETC-point X(110) of the Diagonal Triangle. When L = QL-L9, then X = point on QA-P23.QA-P33. See Ref-34, QFG#1230.
• When L is a line through QA-P4, then QA-Tf8(L) is a line through QA-P2.
 

 
Plaats reactie