National problem · Geometry

Let \(k > 0\). On the sides \(A_1B_1\), \(B_1C_1\) and \(C_1A_1\) of a triangle \(A_1B_1C_1\), points \(C_2\), \(A_2\) and \(B_2\) are chosen, respectively, so that

\frac{A_1C_2}{C_2B_1} = \frac{B_1A_2}{A_2C_1} = \frac{C_1B_2}{B_2A_1} = k .

The construction is then repeated: for every \(i\) with \(2 \leqslant i \leqslant 2012\), points \(C_{i+1}\), \(A_{i+1}\) and \(B_{i+1}\) are chosen on the sides \(A_iB_i\), \(B_iC_i\) and \(C_iA_i\) of the triangle \(A_iB_iC_i\), respectively, so that

\frac{A_iC_{i+1}}{C_{i+1}B_i} = \frac{B_iA_{i+1}}{A_{i+1}C_i} = \frac{C_iB_{i+1}}{B_{i+1}A_i} = \begin{cases} k, & i \equiv 1 \pmod 2, \\[2pt] \dfrac{1}{k}, & i \equiv 0 \pmod 2. \end{cases}

Prove that the lines \(A_1A_{2013}\), \(B_1B_{2013}\) and \(C_1C_{2013}\) meet in one point.