Let $P$ and $Q$ be the projections of $L$ and $N$ onto the line $BC$, respectively. Let $a = BC$, $b = CA$, $c = AB$, let $p$ and $S$ be the half-perimeter and the area of the triangle $ABC$, respectively. Let $X$, $Y$ be the intersection points of the lines $LN$, $MK$ and $BC$, respectively. Without loss of generality we assume that $X$, $Y$ lie on the segment $BC$ (all other positions of these points can be considered in similar way).

It is well known that the radius of the excircle with the center lying opposite to the vertex $C$ is equal to $S/(p-c)$, so $LP = S/(p-c)$. By condition, $AC_1 = BC_1$ and since $N$ is symmetric to $C$ about $C_1$, we have $NC_1 = C_1C$, so $ANBC$ is a parallelogram. If $AH$ is the altitude of $ABC$, then $NQ = AH = 2S/a$. Since $CP = p$, $CB = a$, we have $BP = p-a$. From the equality of the triangles $CHA$ and $BQN$ it follows that $BQ = CH = b \cos \angle C$. Since $LP \parallel NQ$, we see that $\triangle XLP \sim \triangle XNQ$. Then $\frac{XP}{XQ} = \frac{LP}{NQ}$. Let $BX = x$, then

$$\frac{x+p-a}{x+b \cos \angle C} = \frac{S/(p-c)}{2S/a} \Rightarrow x = \frac{2(p-c)(p-a) - ab \cos \angle C}{c-b}$$

Similarly, if $y = CY$, then $y = \frac{2(p-b)(p-a) - ac \cos \angle B}{b-c}$. To prove the required statement, it suffices to show that $x+y=a$. Indeed,

$$\begin{align*} x+y &= \frac{2(p-c)(p-a) - ab \cos \angle C}{c-b} + \frac{2(p-b)(p-a) - ac \cos \angle B}{b-c} = \\ &= \frac{2(p-c)(p-a) - 2(p-b)(p-a) + ac \cos \angle B - ab \cos \angle C}{c-b} = \\ &= \frac{2(b-c)(p-a) + ac \cdot \frac{a^2+c^2-b^2}{2ac} - ab \cdot \frac{a^2+b^2-c^2}{2ab}}{c-b} = \\ &= \frac{(b-c)(b+c-a) + c^2-b^2}{c-b} = \frac{a(c-b)}{c-b} = a. \end{align*}$$