Question on a collinear configuration on quadrilateral

First let us state and prove a lemma:

In the figure, if $AM=MD$, $HP=PJ$ and $HJ || AD$,

then $N, P, M$ are collinear.

Proof:

Consider $\Delta NHP$ and $\Delta NAM$,

$\angle NHP= \angle NAM$ and $\frac{NH}{HP}=\frac{NA}{AM} \implies \Delta NHP \sim \Delta NAM.$

$\therefore \angle HNP= \angle ANM$.

$\therefore N, P, M$ are collinear. $$$$ $\mathbf{Main \; proof:}$

Let us redraw the figure as follows:

$\mathbf{Construction:}$

In the figure, $EH \; || \; BN,$ $HJ \; || \; AD$ and $ \; JF_1|| \; NC$.

$EF_1$ intersects $HJ$ at $P_1$

$$$$ $\mathbf{Proof:}$

$\mathbf{Part \; (i):} $ We first prove that $F_1=F$ and $P_1=P$. $$$$ (1) $\Delta ANB \sim \Delta AHE \implies \frac{AE}{EB}=\frac{AH}{HN}$

(2) $\Delta NAD \sim \Delta NHJ \implies \frac{AH}{HN}=\frac{DJ}{JN}$

(3) $\Delta DNC \sim \Delta DJF_1 \implies \frac{DJ}{JN}=\frac{DF_1}{F_1C}$

(4) Hence $\frac{AE}{EB}=\frac{DF_1}{F_1C}$.

(5) Thus $F_1=F.$

(6) Note that $\Delta F_1PJ \cong \Delta EHP_1 \implies F_1P_1=EP_1.$

(7) Hence $P_1=P$

$$$$ $\mathbf{Part \; (ii):}$

(8) Note that from $\Delta F_1P_1J \cong \Delta EHP_1, P_1J=HP_1 $, i.e. $P_1$ is the mid-point of $HJ$.

(9) From our lemma, $N, P_1, M$ are collinear.

(10) Since $P_1=P$, $N, P, M$ are collinear.