**QL-2P4:**

**Orthocentric Pedal Quadrangle Centers**

*Orthocentric Pedal Quadrangles*

For a quadrigon the pedal quadrangle of the Isogonal Center QA-P4 is a parallelogram; for a quadrilateral the pedal quadrangle of the Miquel Point QL-P1 degenerates collinear on QL-L3. Here for a quadrilateral two points are described, whose pedal quadrangle are orthocentric. These points – without their property – are already mentioned by Clawson. See Ref-22, page 248 (38).

Note: a Quadrangle is Orthocentric when the Orthocenter of each Component Triangle coincide with the 4

^{th}point of the Quadrangle.Clawson describes two points X, Y on the “circumcentric circle” (Miquel Circle QL-Ci3) as common intersections of circles through Si,j and Sk,l, orthogonal to the circumcircle through the Miquel Point QL-P1 and Si,j and Sk,l (Si,j intersection of Li and Lj). These points are QL-2P4a and QL-2P4b and have orthocentric pedal quadrangles.

See also Ref-34, EQF-messages #429, #430, #431.

*Coordinates:*The coordinates of QL-2P4a/b are very complicated and therefore not mentioned here.

However the coefficients of the line through QL-2P4a/b are relatively simple.

*1*

^{st}CT-coefficient QL-2P4a.QL-2P4b:b

^{2}c^{2}l (m - n) (-2 a^{2}l + a^{2}m + b^{2}m - c^{2}m + a^{2}n - b^{2}n + c^{2}n) * (-a

^{2}b^{2}l + b^{4}l - a^{2}c^{2}l - 2 b^{2}c^{2}l + c^{4}l + a^{2}b^{2}m - b^{4}m + b^{2}c^{2}m + a^{2}c^{2}n + b^{2}c^{2}n - c^{4}n)

*Properties:*- QL-2P4a/b are the Clawson-Schmidt Conjugates (QL-Tf1) of QL-2P1a/b.
- QL-2P4a.QL-2P4b is the Perpendicular Bisector of QL-P1.CSC(QL-P5), where CSC = Clawson-Schmidt Conjugate QL-Tf1.
- QL-2P4a and QL-2P4b are collinear with CSCe(QL-L1) and CSCe(QL-P3.QL-P4), where CSCe (Line) = QL-Tf3 = Center of the Clawson Schmidt Conjugate of "Line" (since the Clawson Schmidt Conjugate of a Line is a Circle, there is a center).

Here CSCe(QL-P3.QL-P4) is a point on the Steiner Line QL-L2

and CSCe(Newton Line) is the Reflection of QL-P1 in QL-2P4a.QL-2P4b. - The midpoint of QL-2P4a and QL-2P4b lies on the line CSC(QL-P5).CSC(QL-P7).
- Let L be a parallel to QL-P3.QL-P4 through QL-P1, let L' be the reflection of L in the 1st Steiner Axis, then the QL-Line Isoconjugate QL-Tf2 of L' is QL-2P4a.QL-2P4b. See [34], EQF-message #431.