Hello friends here is my Generalisation of Napoleon Configuration and Morley Configuration

Morey Generalisation

Pure Generalisation of Napoleon Configuration

Sorry Sir Glorfindel, as I was new on this channel so I am not much experience about how to use it. I will re write the generalisation here and important line that will be helpful.

Napoleon configuration generalisation statement : Let there be six arbitrary point A1,A2,....,A6 and let the apex of Equilateral triangle made on base {A1,A2},....{A5,A6} be {B1,B2,....,B6} then G(B1,B2,B3) ; G(B3,B4,B5);G(B5,B6,B1) makes equilateral triangle and when the points A3 and A4 will coincide and cyclically, point (A5,A6) and (A1,A2) coincides then we will get Napoleon configuration . So we can say that Napoleon Configuration is very special case of this configuration & This will valid also when all the six equilateral triangle are drawn inwardly .(You can see more regarding in above link ).

Angle Trisector theorem or Generalisation of Morley configuration: Link

Description___________ THE ANGLE TRISECTOR and Bisector have a very long history . The angle bisector theorem appears as Proposition 03 of Book VI of Euclid elements . Before talking about Angle trisection ,we first look for Angle Bisector (See Figure 01). Take two line segment A1A3 , A2A4 intersect at O . Let the other two lines B1B3 & B2B4 be it’s angle bisector such that ∠A1OB2=∠B2OA2& ∠A2OB3= ∠B3OA3 and when { A1,A2,.....,B1,B2,.....,B4} lies on circle then { A1A2A3A4} and {B1B2B3B4} makes Rectangle . This result may not be new and can be derived but we explain this because we are doing similar thing on Angle trisector. Figure 1.0 Now Angle Trisector is a classical problem and both famous Mathematician Euclid and Archimedes have try to work on them . The Angle Trisection is used in Morley & it is one of the mysterious topic . Here is the description of our observations (see figure 2.0) . Figure 2.0 Let ( A1A3), (A2A4), (B1B5), (B2B6) , (B3B7) ;(B4B8) be line segment such that {A1,A2,A3,A4,B1,.....,B8) lies on the circle such that ∠A1OB1=x ∠B1OB5=x ∠B2OA2=x ∠A4OB6=x ∠B6OB5=x ∠B5OB3=x ∠A2OB3=60-x ∠B3OB4=60-x ∠B4OA3=60-x ∠A4OB7=60-x ∠B7OB8=60-x ∠B8OB1=60-x Or in simple words we can say that line B1B5,B2B6,B3B7 , B4B8 acts as a angle Trisector of line A1A3& A2A4 which intersect at O. Then following Results always holds true which is independent of the variable X. [1]m(B5,O) ; m( B4,B6) ; (B3,B6) makes equilateral triangle [2] m(B3,B6); m( B1,B4);m( B3,B4) makes equilateral triangle. [3] m(B4,B1); m(B3,B1) ; m( B2,O) makes equilateral triangle. [4] m(B6,O) ; m(B5,B7);m( B5,B8) makes equilateral triangle. [5] m(B5,B8) ; m( B7,B8);m( B2,B7) makes equilateral triangle. [6] m(B2,B7) ; m(B2,B8); m( B1,O) makes equilateral triangle & m( B4,B7) ;m(O,B5); m(O,B6) makes equilateral triangle Similarly we can observe m(B3,B8); m(O,B1); m( O,B2)makes equilateral triangle. [7] m{B6,m(B5,O)} ; m{B5,m(B6,O)} ;m{m(B4,B6);m(B5,B7)} makes equilateral triangle. [8] (A1A4),(A2A3),(B7B8), (B3B4),are parallel & Similarly (B1,B2);(A1,A2); (A3,A4) ; (B5,B6) are parallel. [9] m(B4,B7); m( B2,B5) ;m(B3,B6) &m( B4,B7); m( B1,B6); m(B5,B8) makes equilateral triangle. [10] m(B2,B5); m(B3,B8); m(B1,B4) &m(B3,B8) ; m( B1,B6); m(B2,B7) makes 2 equilateral triangle (see Figure 2.0).

Best regards Jayendra Jha and Sankalp Savarn