So, to get this completed and fully define the point C, we have to write a circle with center M1 and radius M1H2 and its intersection with the extension of the segment L1M1 is the point C (and B too).
Dear colleague Nikolaos, thank you for your interest in this problem!
Your solution, in my opinion, is almost perfect.
I suggest that once you obtain point C, connect it to point H2 and draw a straight line CH2. From point L1, draw a perpendicular line L1D to the line CH2 and create a circle with a radius of L1D. Draw a tangent line from point B to the circle with a radius of L1D. The tangent will intersect the line CH2 at point A. The triangle ABC is then constructed.
As above, it is easy to get B & C. Once we have B, C & L_1, one can see that the locus of all points P such that PL_1 is the bisector of angle CPB is a circle with center on CB and of radius determined by CL_1 & BL_1. Then A is at the intersection of this circle and CH_2. I may be wrong but I believe there is a second point A (since the circle and the line will most likely intersect twice).
Gabriel T. Prajitura Dear colleague, English is not my native language, and therefore, I sometimes have to use a translator. Perhaps it's letting me down this time.
Could you describe the construction steps in more detail?
Please write in more detail how you obtain points B and C.
Point A is the only one, an error in construction.
Gabriel T. Prajitura I apologize; I accidentally missed your previous response. Thank you for taking the time to share your version.
In my view, the correct solution appears as follows:
After obtaining points B and C, connect points C and H_2. Line CH_2 passes through point A. From point L_1, drop a perpendicular to line CH_2 – this will be the radius of the semicircle inscribed in triangle ABC, and the diameter lying on side BC. From point B, draw a tangent to this semicircle. The tangent will intersect line CH_2 at the unique point A.