Nice problem!I think it must can be solved by just area chasing,but I just find this solution which is not beautiful. Let $AMCK$ and $ABCJ$ are parallelograms,easy to know that $\vartriangle ABM\cong\vartriangle CJK$ and $\vartriangle ABK\cong\vartriangle CJM$.Let $AB$ meets $CN$ at $H$,meets $CM$ at $I$,$AJ$ meets $CM$ at $G$. $\frac{IH}{IB}=\frac{IH}{IA}\cdot\frac{IA}{IB}=\frac{IC}{IM}\cdot\frac{IG}{IC}=\frac{IG}{IM}$,so $GH\parallel MB$,so $GH\parallel DN\parallel JK$. $S_\vartriangle ABN=S_\vartriangle ABK\cdot\frac{HN}{HK}=S_\vartriangle CJM\cdot\frac{GD}{GJ}=S_\vartriangle CDM$.$\Box$

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I'm not sure of any other good synthetic way. But, the problem can be done using Cartesian coordinate system and the area formula for a triangle. More precisely, if we assume $A=(a,y_1)$, $B=(b,y_2)$, $C=(c,y_2)$, $D=(d,y_1)$ and $M=(m_1,m_2)$, $N=(n_1,n_2)$ (we are assuming $AD$ is parallel to $x$-axis), then we should have \[ \frac{m_2-y_1}{m_1-a} = \frac{y_2-n_2}{c-n_1} \quad \text{and} \quad \frac{n_2-y_1}{n_1-d} = \frac{y_2-m_2}{b-m_1} \quad \quad (1) \]as $AM\parallel CN$ and $BM\parallel DN$. We only need to show that \[ \frac{1}{2} \begin{vmatrix} 1 & 1 & 1\\ a & b & n_1 \\ y_1 & y_2 & n_2 \end{vmatrix} = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1\\ c & d & m_1 \\ y_2 & y_1 & m_2 \end{vmatrix} \quad \quad (2) \]as they are areas of triangles $ABN$ (left one) and $CDM$ (right one). It can be shown that $(1)$ implies $(2)$ using some algebraic manipulations.

Does anyone have pure geometry solution？

Define $X=AM\cap DN$ and $Y=BM\cap CN$. Let $\angle MYN=\theta$ and let $\lambda=\frac{1}{2}\sin(\theta)$. $$[ABN]=[AYN]+[NYB]+[BYA]=\lambda(YM\cdot YN+YB\cdot YN+YB\cdot AM)=\lambda(YM\cdot YN+YB\cdot AX)$$Similarly $[CDM]=\lambda(XM\cdot XN+XD\cdot CY)$. We are done as $YM\cdot YN=XM\cdot XN$ and $YB\cdot AX=XD\cdot CY$ (because $\Delta AXD\sim\Delta CYB$). Remark: An alternative approach would be to send parallelogram $MYNX$ to a unit square through an affine transformation.

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Use barycentric coordinates with respect to $ABC$. Then $D=(1,t,-t)$ for some $t\in \mathbb{R}$. Let $M=(x,y,z)$. Then $P_{\infty AM}=(-y-z:y:z)$ and $P_{\infty BM}=(x:-x-z:z)$, so $N=(-y-z:y:s)$ for some $s\in \mathbb{R}$ (because $AM\parallel CN$) and $N\in DP_{\infty BM}$ (because $BM\parallel DN$), hence $N=\left (-y-z:y:z\frac{tx-y}{(t+1)x+z}\right )=(-(y+z)((t+1)x+z):y((t+1)x+z):z(tx-y))$. Then using $x+y+z=1$ we get $N=\left (\frac{(y+z)((t+1)x+z)}{z},-\frac{y((t+1)x+z)}{z},y-tx\right )$. Hence$$\frac{[ABM]}{[ABC]}=\begin{vmatrix}1&0&0\\0&1&0\\\frac{(y+z)((t+1)x+z)}{z}&-\frac{y((t+1)x+z)}{z}&y-tx\end{vmatrix}=y-tx$$and$$\frac{[CDM]}{[ABC]}=\begin{vmatrix}0&0&1\\1&t&-t\\x&y&z\end{vmatrix}=y-tx.$$Therefore, $[ABM]=[CDM]$, as desired.

The result is valid though $M, N$ are outside or inside the specified polygon. Denotes the signed area of the polygon $X$ as $[X]$ Lemma Let $ABCD$ be a trapezoid s.t. $CD\parallel AB$ ; $E $ be a point then $[AECD]=[BECD]$ the proof is easy. Applying the lemma twice we get $[BAND]=[CAND ]=[CMND ]$ but $[BAND]=[BAN ]+[BND ]$ and $[CMND]=[DCM ]+[DMN ] $ besides $[BND ]=[DMN ](\because DN\parallel BM)$ we deduces then $[BAN ]=[DCM ]$ best regards. RH HAS