Triangle problem (open)

[ins-]

It is given a triangle ABC. Inside triangle are chosen two points M and N. If <MAC=<NBC, <MCA=<NCB and AN=BM is it true that AC=BC?

[mateo]

It's not true. Let's <MAC=a, <MCA=b. If true 2*(cos(a))^2*(tg(a)+tg(b))=1, we anable to perform geometrical construction.

[ins-]

Thank you very much for the answer. Can you be more detailed?

[terminator-II]

"open problem" such a claim imposes some obligation

[mateo]

If we consider rectangular coordinate system with axe OX == AC (right-handed system) and OX == BC (left-handed system), we can get that

BM^2 = AB^2 * ((cos(A) - (sin(B)/sin(C)) / (tg(a) + tg(b)))^2 + (sin(A) - (sin(B)/sin(C)) * (tg(a)/ (tg(a) + tg(b)))^2)

AN^2 = AB^2 * ((cos(B) - (sin(A)/sin(C)) / (tg(a) + tg(b)))^2 + (sin(B) - (sin(A)/sin(C)) * (tg(a)/ (tg(a) + tg(b)))^2)

If we transform AN = BM, we can get that 2 * (cos(a))^2 * (tg(a)+tg(b)) = 1 or <A = <B for execution equality.

[ins-]

I'm sorry for the stupid questions but I have 3 more questions:
1. why are you using coordinate system (Analytical Geometry) isn't it possible the problem to be solved by using only trygonometry?
2. Are A,B,C equal to <BAC, <ABC, <BCA respectively and <MAC=a, <MCA=b?
3. Can you describe how you obtained your first equality?

[ins-]

I'm sorry for loosing your time. The problem is very easy. I solved it in my own very simple way. It requires more concentration. No coordinate system is required. Only simple trigonometry.