We have a triangle $\Delta ABC$ ($AC$ is not equal to $BC$ and thus the triangle is not isosceles) with angle bisectors $AA_1$, $BB_1$ and $CC_1$ ($A_1$, $B_1$ and $C_1$ are on the sides $BC$, $AC$ and $AB$ respectively). If angle $AA_1C_1$ is equal to angle $BB_1C_1$, find out angle $BCA$.
I know that the answer is 120 degrees but I couldn't prove much.
This problem was given to me by my maths teacher. He gave me the following hint: let point X lie on $CB_1$ and $CX=CA_1$. After some time i finally managed to solve the task. Here is the solution:
Without loss of generality we can have $AC>BC$ ($\Delta ABC$ is not isosceles). It's easy to prove that $CB_1 > CA_1$, so point X lies on the $segment$ $CB_1$. Next we show that $\Delta CIX$ is congruent to $\Delta CIA_1$ and therefore $\Delta C_1IX$ is congruent to $\Delta C_1IA_1$. From this we have angle $C_1XI$ is equal to angle $IA_1C_1$ which is equal to angle $IB_1C_1$ (from the statement). Now we can prove that quadrilateral $B_1C_1IX$ is inscribed. $=>$ (angle $XB_1C_1$ $+$ angle $XIC_1$) is equal to $180$ degrees. By using simple angle expressions we find that angle $IB_1C_1$ is equal to $\gamma$ $-$ $90$ degrees.
After knowing this by making more angle expressions we find that angle $AC_1B_1$ is equal to $90$ degrees $- \alpha - \beta/2$ and angle $B_1C_1C$ is equal to $\alpha/2 + \beta$. Now we use three sinus theorems and get this equality: $AB_1/B_1C=(sin(\gamma/2)/sin(\alpha))*(cos(\alpha+\beta/2)/sin(\alpha/2+\beta))$. Now using the fact that $BB_1$ is angle bisector in $\Delta ABC$ and another sinus theorem for $\Delta ABC$ we obtain only trigonometric equality: $sin(\gamma)/sin(\alpha)=(sin(\gamma/2)/sin(\alpha))*(cos(\alpha+\beta/2)/sin(\alpha/2+\beta))$. After some simplification we have this: $2*cos(\gamma/2)=sin(\gamma/2-\alpha/2)/sin(\gamma+\alpha/2)$. Now we use sum and difference trigonometric identities and the equality looks like this: $2*sin(\gamma)*cos(\alpha/2)+2*cos(\gamma)*sin(\alpha/2)=tg(\gamma/2)*cos(\alpha/2)-sin(\alpha/2)$. We make some groping and then some groping and use a half angle trigonometry formula and we obtain this: $sin(\alpha/2)*(2*cos(\gamma)+1)-tg(\gamma/2)*cos(\alpha/2)*(1-2*(1+cos(\gamma))=0$. Now $2*cos(\gamma)+1$ is a common multiple and either he is $0$ or $sin(\alpha/2)+tg(\gamma/2)*cos(\alpha/2)$ because the product is $0$ if one of the multiples is $0$. The second case is impossible due to $\alpha$ and $\gamma$ being angles in triangle and the trigonometric functions from their halves are always positive. And from the first case we finally found that $\gamma$ is $120$ degrees.
If something in the solution isn't clear, feel free to ask me. I summarized some moments because this text would've been two times bigger.
P.S. I hope that i haven't done any mistakes.
