i can solved you dear

Budee U Zaman,

Show your solution here

Thats how you find the center

Of mass of a cardboard triangle?

On a straight line you take the point H2 and a perpendicular line on it.

Write the segment L1M1 and extend it. It will intersect the straight line (of point H2) at point B. Then you have BM1

Extend BM1 for an equal segment and you have the point C.

Draw a perpendicular line from the point L1 to the segment H2B and name the intersection e.g. D.

Write a circle with center L1 and radius L1D.

From the point C draw a tangent to this circle and it will intersect the extension of the segment H2B at point A. So you have the triangle ABC.

The above uploaded is not complete.

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).

Nikolaos Vrettos

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.

Thank you dear Liudmyla.

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.

Same way as Nikolaos Vrettos. M_1 is the midpoint of the hypotenuse in the

right triangle BCH_2, therefore points B & C are on the line M_1L_1, at distance M_1H_2 from M_1.

I don't know if there are 2 or one possible points A unless I check the computation using the equation of the locus, which I did not have time to.

yes

Gabriel T. Prajitura You didn't use point M1, which is given in the problem statement. This is a hint. Point P is not specified in the statement.

Why don't you read my answer. You can see there M_1 at least 4 times. How did I not use it?

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.