Morley defines a 2nd circle with radius 1/(n-1) radius of the centric circle (1/2 for the triangle and 1/3 for the quadrilateral, etc.). The location of the center of this second circle is defined as the Ratiopoint nL-n-P4.nL-n-P3 (1 : n-1).
nL-n-P5 can be constructed in a recursive way:
• Construct the perpendiculars of the n versions of (n-1)L-n-P5 of the Component (n-1)-Lines to the omitted line. They will concur in nL-n-P4.
• Construct nL-n-P5 = nL-n-P4.nL-n-P3 (1 : n-1).
• Construct the perpendiculars of the n versions of (n-1)L-n-P5 of the Component (n-1)-Lines to the omitted line. They will concur in nL-n-P4.
• Construct nL-n-P5 = nL-n-P4.nL-n-P3 (1 : n-1).
For the triangle, the 2nd circle is the Euler Circle (or Nine-point Circle or Feuerbach Circle) with center the Nine-point Center X(5) being 3L-n-P4.3L-n-P3 (1:1).
The 4 perpendiculars drawn from the 4 points of the Component 3-Lines to the 4th line concur in the 2nd Orthocenter of the 4-Line being 4L-n-P4. The center of the 2nd circle in the 4-Line will be at 1/3 on the segment 4L-n-P4.4L-n-P3 of the 4-Line.
The 5 perpendiculars drawn from the 5 points of the Component 4-Lines to the 5th line concur in the 2nd Orthocenter of the 5-Line being 5L-n-P4. The center of the 2nd circle in the 5-Line will be at 1/4 on the segment 5L-n-P4.5L-n-P3 of the 4-Line, etc.
The 4 perpendiculars drawn from the 4 points of the Component 3-Lines to the 4th line concur in the 2nd Orthocenter of the 4-Line being 4L-n-P4. The center of the 2nd circle in the 4-Line will be at 1/3 on the segment 4L-n-P4.4L-n-P3 of the 4-Line.
The 5 perpendiculars drawn from the 5 points of the Component 4-Lines to the 5th line concur in the 2nd Orthocenter of the 5-Line being 5L-n-P4. The center of the 2nd circle in the 5-Line will be at 1/4 on the segment 5L-n-P4.5L-n-P3 of the 4-Line, etc.
