Let the lines $KL$ and $NP$, $LJ$ and $PM$, $JK$ and $MN$ meet at points $Q$, $R$, $S$, respectively. By Desargues' theorem on perspective triangles, the lines $JM$, $KN$ and $LP$ are concurrent if and only if the points $Q$, $R$ and $S$ are collinear.

Without loss of generality, we may assume that $X$ lies on the arc $FD$ of the incircle that does not contain $E$. Consequently, $K$ lies on the side $FD$, while $L$ and $J$ lie on the respective extensions of the sides $DE$ and $EF$.

Consider the cyclic quadrangle $DEFX$: The diagonals meet at $K$, and the extensions of the opposite sides meet at $L$ and $J$, respectively, so the line $LJ$ is the polar of $K$ with respect to the incircle – in what follows, all polar lines are considered with respect to the incircle. Since the line $FD$ is the polar of $B$, and $K$ lies on the line $FD$, it follows that $B$ lies on the line $LJ$. Similarly, $C$ lies on the line $JK$, and $A$ lies on the line $KL$. Consequently, the lines $AQ$ and $CS$ meet at $K$.

Projectively, the lines $FD$ and $LJ$ meet at some point $T$. Notice that $T$ lies on the polar lines of $B$ and $K$ to deduce that the line $BK$ is the polar of $T$, so the cross-ratio $(TFKD)$ is harmonic. Let further the lines $BK$ and $PM$ meet at $U$. Read from $B$, the cross-ratio $(RPUM)$ equals $(TFKD)$, so it is harmonic; that is, $U$ is the harmonic conjugate of $R$ relative to $M$ and $N$. Consequently, the points $Q$, $R$ and $S$ are collinear if and only if the lines $MQ$, $NU$ and $PS$ are concurrent.

To prove the lines $MQ$, $NU$ and $PS$ concurrent, simply check that
$$\frac{QN}{QP} \cdot \frac{UP}{UM} \cdot \frac{SM}{SN} = 1.$$