First show that the line $LK$ is parallel to the side $AB$. Indeed, since $C_1L$ and $C_1K$ are the bisectrices of the angles $AC_1C$ and $BC_1C$ respectively we have
$$\frac{AL}{LC} = \frac{AC_1}{CC_1} = [AC_1 = BC_1] = \frac{C_1B}{CC_1} = \frac{BK}{KC}.$$

So $LK \parallel AB$. Since $A_1$ and $B_1$ are the midpoints of $BC$ and $AC$ respectively, we have $B_1A_1 \parallel AB \parallel LK$. Therefore
$$\frac{B_1A_2}{B_1C_1} = \frac{A_1B_2}{A_1C_1}. \qquad (1)$$
Since $LA_2 \parallel AC_1$ we have $\frac{LA_2}{AC_1} = \frac{B_1A_2}{B_1C_1}$, so $LA_2 = \frac{B_1A_2}{B_1C_1} AC_1$.
Similarly, $KB_2 = \frac{A_1B_2}{A_1C_1} C_1B$.
Since $AC_1 = C_1B$, it follows from (1) that $LA_2 = KB_2$.
Let $CC_1$ meets $LK$ at $C_2$. Since $CC_1$ is the median of $ABC$ and $LK \parallel AB$, we have $LC_2 = C_2K$. Therefore,
$$|A_2C_2| = |LC_2 - LA_2| = |KC_2 - B_2K| = |C_2B_2|.$$
Thus the line $C_1C_2$ passes through the midpoints $C_1$ and $C_2$ of the bases of the trapezoid $AA_2B_2B$. It is well-known fact that the line $C_1C_2$ and the extensions of the sides $AA_2$, $BB_2$ of the trapezoid $AA_2B_2B$ are concurrent.