Let the circle $\omega$ inscribed in a hexagon $ABCDEF$ tangents the sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$ at the points $A_4$, $A_3$, $A_2$, $A_1$, $A_6$, $A_5$ respectively.

(1) Let $A_6A_2$ be the polar line of the point $L$ with respect to $\omega$ and let $A_3A_5$ be the polar line of the point $N$ with respect to $\omega$.

Let $l_X$ denote the polar line of the point $X$. If $A_6A_2 \cap A_3A_5 = X$, then $X \in l_N$; $X \in l_L$. Here we used the known result from projective geometry that if $X \in l_Y$, then $Y \in l_X$. Hence the polar line of $X$ is $l_X = LN$.

(2) Let $A_5A_1$ and $A_4A_2$ be the polar lines of $M$ and $K$ respectively. Let $A_5A_1 \cap A_4A_2 = Y$. The polar line of $Y$ is $l_Y = MK$.

(3) Let $A_4A_3$ and $A_6A_1$ be the polar lines of $B$ and $E$ respectively. Let $A_4A_3 \cap A_6A_1 = Z$. The polar line of $Z$ is $l_Z = BE$.

By Pascal's theorem the points $A_6A_2 \cap A_3A_5 = X$, $A_2A_4 \cap A_5A_1 = Y$ and $A_4A_3 \cap A_1A_6 = Z$ are collinear. Suppose that the points $X$, $Y$, $Z$ lie on the line $m$. Then the pole of $m$ lies on the lines $l_X$, $l_Y$, $l_Z$. Therefore the lines $l_X$, $l_Y$ and $l_Z$ are concurrent. Thus the lines $LN$, $MK$ and $BE$ are concurrent.