QA-Tr1: QA-Diagonal Triangle
Let P1, P2, P3, P4 be the defining Quadrangle Points.
Let S1 = P1.P2 ^ P3.P4, S2 = P1.P3 ^ P2.P4 and S3 = P1.P4 ^ P2.P3.
Now S1.S2.S3 is the QA-Diagonal Triangle of the Reference Quadrangle.
Areas:
Area QA-Diagonal Triangle in CT-notation:
2 p q r Δ / ((p + q) (p + r) (q + r))
Area QA-Diagonal Triangle in DT-notation:
2 S
Properties:
- Every Component Triangle Pi.Pj.Pk is the Anticevian Triangle of the 4thpoint Pl (for all combinations of (i,j,k,l) ∈ (1,2,3,4)).
- The vertices of the QA-Diagonal Triangle combined with the Isotomic Center form a new quadrangle with the same QA-Centroid as the Reference Quadrangle.
- The Isogonal Center of the Quadrangle S1.S2.S3.QA-P4 is the Involutary Conjugate of QA-P4 (note Eckart Schmidt).