Answer is NO. We will find a non-parallelogram $ABCD$ satisfying the given conditions. Let $WXYZ$ be the quadrilateral formed by lines $AK,BL,CM,DN$, more particularly $$ W = \overline{AK} \cap \overline{BL} ~,~ X = \overline{BL} \cap \overline{CM} ~,~ Y = \overline{CM} \cap \overline{DN} ~,~ Z = \overline{DN} \cap \overline{AK} $$The given condition is simply saying $$ \frac{WX}{BL} = \frac{XY}{CM} = \frac{YZ}{DN} = \frac{ZW}{AK} \qquad \qquad (1)$$Denote by $[\lambda]$ the area of figure $\lambda$ ; let $d,a, b, c$ equal $[ABC],[BCD],[CDA],[DAB]$, respectively. Observe that $$a + c = b + d = [ABCD] \qquad \qquad (2)$$ Claim: For any convex quadrilateral $ABCD$, irrespective of its shape we have \begin{align*} 1 - \frac{WX}{BL} = \frac{2 d}{a + 2 c + 2 d} + \frac{b}{2 a + b + 2d} \\ 1 - \frac{XY}{CM} = \frac{2 a}{b + 2 d + 2 a} + \frac{c}{2 b + c + 2 d} \\ 1 - \frac{YZ}{DN} = \frac{2b}{c + 2 a + 2 b} + \frac{d}{2 c + d + 2b} \\ 1 - \frac{ZW}{AK} = \frac{2 c}{d + 2 b + 2 c} + \frac{a}{2 d + a + 2c} \end{align*}Proof: We will only show $$1 - \frac{WX}{BL} = \frac{2 d}{a + 2c + 2 d} + \frac{b}{2a + b + 2 d}$$and other results would follow analogously. Observe that \begin{align*} &\frac{BW}{WL} = \frac{[BWK]}{[LWK]} = \frac{[BWA]}{[LWA]} = \frac{[BWK] + [BWA]}{[LWK] + [LWA] } = \frac{[BAK]}{[LAK]} \\ \implies \frac{BW}{BL} &= \frac{[BAK]}{[BALK]} = \frac{d/2}{ [ABCD] - [DAL] - [CLK]} \\ &= \frac{d/2 }{ (a/2 + b/2 + c/2 + d/2) - b/2 - a/4} = \frac{2 d}{a + 2 c + 2 d} \end{align*}We also have \begin{align*} &\frac{LX}{XB} = \frac{[LXC]}{[BXC]} = \frac{[LXM]}{[BXM]} = \frac{[LXC] + [LXM]}{[BXC] + [BXM]} = \frac{[CLM]}{[CBM]} \\ \implies \frac{LX}{LB} &= \frac{[MLC]}{[BMLC]} = \frac{[MDC]/2}{[ABCD] - [BAM] - [DML]} \\ &= \frac{b/4}{(a/2 + b/2 + c/2 + d/2) - c/2 - b/4} = \frac{b}{2 a + b + 2 d } \end{align*}Finally, note $$1 - \frac{WX}{BL} = \frac{BW}{BL} + \frac{LX}{LB}$$This proves our claim. $\square$ Because of our Claim, $(1)$ is equivalent to \begin{align*} & ~~~~ \frac{2 d}{a + 2 c + 2 d} + \frac{b}{2 a + b + 2d} = \frac{2 a}{b + 2 d + 2 a} + \frac{c}{2 b + c + 2 a} \\ &= \frac{2b}{c + 2 a + 2 b} + \frac{d}{2 c + d + 2b} = \frac{2 c}{d + 2 b + 2 c} + \frac{a}{2 d + a + 2c} \end{align*}We show that for any $a,b,c,d$ satisfying $$ a+c=b+d ~,~ b+a = 2c ~,~ d+c = 2a ~,~ b+2d = 3a ~,~ d+2b = 3c \qquad \qquad (3)$$the above four quantities are equal. Indeed, if $(3)$ is true then \begin{align*} \frac{2d}{a+2c + 2d} + \frac{b}{2a + b + 2d} &= \frac{2d}{a + 2(2a)} + \frac{b}{2a + (3a)} = \frac{2d + b}{5a} = \frac{3}{5} \\ \frac{2a}{b+2d+2a} + \frac{c}{2b + c + 2a} &= \frac{2a}{(3a) + 2a} + \frac{c}{ 2(2c) + c} = \frac{2}{5} + \frac{1}{5} = \frac{3}{5} \\ \frac{2b}{c + 2a + 2b} + \frac{d}{2c + d + 2b} &= \frac{2b}{c + 2(2c)} + \frac{d}{2c + (3c)} = \frac{2b + d}{5c} = \frac{3}{5} \\ \frac{2c}{d + 2b + 2c} + \frac{a}{2d + a + 2c} &= \frac{2c}{(3c) + 2c} + \frac{a}{2(2a) + a} = \frac{2}{5} + \frac{1}{5} = \frac{3}{5} \end{align*}Next we construct a convex quadrilateral $ABCD$ for which $(3)$ is satisfied. Pick two perpendicular lines $\ell_1,\ell_2$, let $T = \ell_1 \cap \ell_2$. Pick points $A,C \in \ell_1$ and $B,D \in \ell_2$ with $T$ lying on segments $AC,BD$ such that $$ TA = 3, TB = 10 , TC = 4, TD = 4 $$Let $p,q,r,s$ denote $[TAB],[TBC],[TCD],[TDA]$, respectively. Observe, \begin{align*} p = \frac{3 \cdot 10}{2} = 15 ~,~ q = \frac{10 \cdot 4}{2} = 20 ~,~ r= \frac{4 \cdot 4}{2} = 8 ~,~ s = \frac{4 \cdot 3}{2} = 6 \\ \implies a = q + r = 28 ~,~ b = r + s = 14 ~,~ c = s +p = 21 ~,~ d = p + q = 35 \end{align*}It isn't hard to verify all equalities in $(3)$ indeed hold true. $\blacksquare$ Motivation: We obtain \begin{align*} \frac{2d}{a + 2c + 2d} + \frac{b}{2a + b + 2d} &= \frac{2b}{c + 2a + 2b} + \frac{d}{2c + d + 2b} \\ \iff \frac{2d}{a+2c+2d} - \frac{2b}{c + 2a + 2b} &= \frac{d}{2c + d + 2b} - \frac{b}{2a + b + 2d} \\ \iff \frac{2 \left( d(c + 2a) - b(a+2c) \right)}{(a+2c + 2d)(c + 2a + 2b)} &= \frac{2 \left( d(a+d) - b(c+b) \right)}{(2c + d + 2b)(2a + b + 2d)} \\ \iff \frac{d(a+b+d) - b(b+c+d)}{(a+2c + 2d)(c + 2a + 2b)} &= \frac{d(a+d) - b(c+b)}{(2c + d + 2b)(2a + b + 2d)} \qquad (\text{using } a+c = b+d) \\ \iff \frac{d(a+d) - b(c+b)}{(a+2c + 2d)(c+2a + 2b)} &= \frac{d(a+d) - b(c+b)}{(2c + d + 2b)(2a + b + 2d)} \qquad \qquad (\spadesuit) \end{align*}We similarly have \begin{align*} \frac{2a}{b + 2d + 2a} + \frac{c}{2b + c + 2a} &= \frac{2c}{d + 2b + 2c} + \frac{a}{2d + a + 2c} \\ \iff \frac{2a}{b + 2d + 2a} - \frac{2c}{d + 2b + 2c} &= \frac{a}{2d + a +2c} - \frac{c}{2b + c + 2a} \\ \iff \frac{2 \left( a(d + 2b) - c(b + 2d) \right)}{(b+2d + 2a)(d + 2b + 2c)} &= \frac{2 \left( a(b + a) - c( d + c) \right)}{(2d + a + 2c)(2b + c + 2a)} \\ \iff \frac{a(a+c+b) - c(a+c+d) }{(b + 2d + 2a)(d + 2b + 2c)} &= \frac{a(b+a) - c(d+c)}{(2d + a + 2c)(2b + c + 2a)} \qquad (\text{using } a+c = b+d) \\ \iff \frac{a(b+a) - c(d+c)}{(b+2d + 2a)(d + 2b + 2c)} &= \frac{a(b+a) - c(d+c)}{(2d + a + 2c)(2b + c + 2a)} \qquad \qquad (\clubsuit) \end{align*}First assume $$ (b+2d + 2a)(d + 2b + 2c) \ne (2d + a + 2c)(2b + c + 2a) $$Then $(\spadesuit)$ and $(\clubsuit)$ would respectively imply \begin{align*} d(a+d) = b(c+b) \qquad \qquad (\diamondsuit) \\ a(b+a) = c(d+c) \qquad \qquad (\heartsuit) \end{align*}Thus we obtain \begin{align*} b(c+b) + a(b+a) &= d(a+d) + c(d+c) \\ \implies a^2 - c^2 &= (d^2 - b^2) + (da + dc) - (ab + bc) = (b+d)(b-d) + d(a+c) - b(a+c) \\ &= (b+d)(b-d) + (d-b)(a+c) = 0 \end{align*}Hence $a=c$. Then $(\heartsuit)$ gives $b+a = d+c$, so $b=d$. As $a+c = b+d$, so $a=b=c=d$. Next assume $$ (b+2d + 2a)(d+2b + 2c) = (2d + a + 2c)(2b + c + 2a) $$As $a+c = b+d$, so $$ (b + 2d + 2a) + (d + 2b + 2c) = (2d + a +2c) + (2b + c + 2a)$$It follows $$\{b + 2d + 2a,d + 2b + 2c\} ~,~ \{2d + a + 2c, 2b + c + 2a\}$$are the set of roots of the same quadratic equation, in particular the multi sets are equal. We have two cases: \textbf{Case 1:} $b+2d + 2a = 2d + a + 2c$ and $d + 2b + 2c = 2b + c + 2a$. These respectively give $$ b+a = 2c ~,~ d+c = 2a $$Now we have \begin{align*} \frac{2d}{a + 2c + 2d} + \frac{b}{2a + b + 2d} &= \frac{2a}{b + 2d + 2a} + \frac{c}{2b + c + 2a} \\ \iff \frac{2d}{a+2c+2d} - \frac{2a}{b+2d+2a} &= \frac{c}{2b + c + 2a} - \frac{b}{2a + b + 2d} \\ \iff \frac{2 \left( d(b + 2d) - a(a+2c) \right) }{(a+2c + 2d)(b+2d + 2a)} &= \frac{2 \left( c(a+d) - b(b+a) \right)}{(2b + c + 2a)(2a + b + 2d)} \\ \iff \frac{d(a+c+d) - a(b+d+c)}{a+2c + 2d} &= \frac{c(a+d) - b(b+a)}{2b + c + 2a} \\ \iff \frac{d(c+d) - a(b+c)}{a + 2(2a))} &= \frac{c(a+d) - b(b+a)}{c + 2(2c)} \\ \iff \frac{d(c+d) - a(b+c)}{a} &= \frac{c(a+d) - b(b+a)}{c} \\ \iff \frac{d(c+d)}{a} - \frac{a(b+c)}{a} &= \frac{c(a+d)}{c} - \frac{b(b+a)}{c} \\ \iff 2d - (b+c) &= (a+d) - 2b \\ \iff d - a = c - b \end{align*}This is always true as $a+c = b+d$. This basically means if $b+a = 2c$ and $a+c = b+d$, then required conditions are satisfied!

@above a much simpler counterexample exists, but ok. Here is my full solution, including the motivation behind how we come up with a counterexample: The answer is $\textbf{No}$. $\textbf{Solution:}$ I wanted to present my thought process and how I came up with the counterexample. In what's presented below I explain this. Let $X=AK\cap BL$, $Y=BL\cap CM$, $Z=CM\cap DN$, $W=DN\cap AK$. Let $\frac{XY}{BL}=\frac{YZ}{CM}=\frac{ZW}{DN}=\frac{WX}{AK}=\lambda$. We'll denote the area of $\triangle A_{1}A_{2}A_{3}$ by $S_{A_{1}A_{2}A_{3}}$ or $S_{\triangle A_{1}A_{2}A_{3}}$ and respectively for quadrilaterals. $\textbf{Step 1:}$ Let $G$ be the centroid of $ABCD$, that is, the center of mass for the four points $A,B,C,D$ and also the intersection of $MK$ and $NL$. Then notice that $G$ lies strictly inside $XYZW$ (because $G=MK\cap NL$), so: $$S_{XYZW}=S_{\triangle XGY}+S_{\triangle YGZ}+S_{\triangle ZGW}+S_{\triangle WGX}=$$$$=\lambda(S_{\triangle BGL}+S_{\triangle CGM}+S_{\triangle DGN}+S_{\triangle AGK})=$$$$=\frac{\lambda}{2}(S_{\triangle BLN}+S_{\triangle CMK}+S_{\triangle DLN}+S_{\triangle AKM})=$$$$=\frac{\lambda}{4}(S_{\triangle ALB}+S_{\triangle BMC}+S_{\triangle CND}+S_{\triangle DKA})=$$$$=\frac{\lambda}{8}\Big((S_{\triangle DAB}+S_{\triangle ABC})+(S_{\triangle ABC}+S_{\triangle BCD})+$$$$+(S_{\triangle BCD}+S_{\triangle CDA})+(S_{\triangle CDA}+S_{\triangle DAB})\Big)=$$$$=\frac{\lambda}{8}\times 4S_{ABCD}=\frac{\lambda}{2}S_{ABCD}$$\[\Longrightarrow S_{XYZW}=\frac{\lambda}{2}S_{ABCD}\quad (1)\]We can also show that: $$S_{\triangle AZW}+S_{\triangle CXY}=\lambda(S_{\triangle AND}+S_{\triangle CLB})=\frac{\lambda}{2}(S_{\triangle ABD}+S_{\triangle BCD})=\frac{\lambda}{2}S_{ABCD}$$\[\Longrightarrow S_{\triangle AZW}+S_{\triangle CXY}=\frac{\lambda}{2}S_{ABCD}\quad (2)\]$$S_{\triangle BXW}+S_{\triangle DYZ}=\lambda(S_{\triangle ABK}+S_{\triangle CDM})=\frac{\lambda}{2}(S_{\triangle ABC}+S_{\triangle CDA})=\frac{\lambda}{2}S_{ABCD}$$\[\Longrightarrow S_{\triangle BXW}+S_{\triangle DYZ}=\frac{\lambda}{2}S_{ABCD}\quad (3)\]Also, because $S_{AKCM}=S_{\triangle AKC}+S_{\triangle CMA}=\frac{1}{2}(S_{\triangle ABC}+S_{\triangle CDA})=\frac{1}{2}S_{ABCD}$ and $S_{DNBL}=S_{\triangle DNB}+S_{\triangle BLD}=\frac{1}{2}(S_{\triangle DAB}+S_{\triangle BCD})=\frac{1}{2}S_{ABCD}$ we have that: $$S_{\triangle BXC}+S_{\triangle AZD}=2(S_{\triangle CXK}+S_{\triangle AZM})=$$$$=2(S_{AKCM}-S_{XYZW}-(S_{\triangle AZW}+S_{\triangle CXY}))\overset{(2)}{=}$$$$\overset{(2)}{=}2\left(\frac{1}{2}S_{ABCD}-\frac{\lambda}{2}S_{ABCD}-\frac{\lambda}{2}S_{ABCD}\right)=(1-2\lambda)S_{ABCD}$$\[\Longrightarrow S_{\triangle BXC}+S_{\triangle AZD}=(1-2\lambda)S_{ABCD}\quad (4)\]$$S_{\triangle DYC}+S_{\triangle AWB}=2(S_{\triangle DLY}+S_{\triangle BNG})=2(S_{BLDN}-S_{XYZW}-(S_{\triangle BXW}+S_{\triangle DYZ})\overset{(3)}{=}$$$$\overset{(3)}{=}2(\frac{1}{2}S_{ABCD}-\frac{\lambda}{2}S_{ABCD}-\frac{\lambda}{2}S_{ABCD})=(1-2\lambda)S_{ABCD}$$\[\Longrightarrow S_{\triangle DYC}+S_{\triangle AWB}=(1-2\lambda)S_{ABCD}\quad (5)\]Now, using $(1),(2),(3),(4),(5)$ we can evaluate $\lambda$: $$S_{ABCD}=S_{XYZW}+(S_{\triangle BXW}+S_{\triangle DYZ})+(S_{\triangle AZW}+S_{\triangle CXY})+$$$$+(S_{\triangle CXB}+S_{\triangle DZA})+(S_{\triangle CYD}+S_{\triangle AGB})=$$$$=\frac{\lambda}{2}S_{ABCD}+\frac{\lambda}{2}S_{ABCD}+\frac{\lambda}{2}S_{ABCD}+(1-2\lambda)S_{ABCD}+(1-2\lambda)S_{ABCD}=$$$$=\left(2-\frac{5}{2}\lambda\right)S_{ABCD}\Longrightarrow \boxed{\lambda=\frac{2}{5}}$$ From here on we'll use the results above with $\lambda=\frac{2}{5}$ as well as use the ratios, which $\lambda$ represents. $\textbf{Step 2:}$ Now look at $AKCM$ and let $\frac{AW}{AK}=a<1-\lambda=\frac{3}{5}$ and $\frac{CY}{CM}=b<1-\lambda=\frac{3}{5}$. Then we have that: $$\frac{2}{5}S_{AKCM}=\frac{1}{5}S_{ABCD}=S_{XYZW}=S_{\triangle XYW}+S_{\triangle YZW}=\frac{2}{5}S_{\triangle AYK}+\frac{2}{5}S_{\triangle CWM}=$$$$=\frac{2}{5}\left(bS_{AMK}+(1-b)S_{ACK}+aS_{CKM}+(1-a)S_{AMC}\right)=$$$$=\frac{2}{5}\left(bS_{AMCK}-bS_{CMK}+(1-b)S_{AKCM}-(1-b)S_{AMC}+aS_{CKM}+(1-a)S_{AMC}\right)=$$$$=\frac{2}{5}S_{AKCM}+\frac{2}{5}(a-b)(S_{\triangle CKM}-S_{\triangle AMC})$$$$\Longrightarrow \frac{2}{5}(a-b)(S_{\triangle CKM}-S_{\triangle AMC})=0$$There are two cases: $a=b$ or $S_{\triangle CKM}=S_{\triangle AMC}$ and in the second case that'd mean that $CM\parallel AK$ as $\text{dist}(A,MC)\cdot MC=2S_{\triangle AMC}=2S_{\triangle CKM}=\text{dist}(K,MC)\cdot MS\Longrightarrow \text{dist}(A,MC)=\text{dist}(K,MC)\Longrightarrow AK\parallel CM$. An analogical argument can be said about the quadrilateral $BLDN$. Now we just have to examine the cases. $\textbf{Step 3}$ If $AK\parallel CM$ and $BL\parallel DN$, then $XYZW$ is a prallelogram and so $AK=\frac{5}{2}XW=\frac{5}{2}YZ=CM$ and analogically $DN=BL$. But then $AK\parallel CM$ and $AK=CM$, so $AKCM$ is a parallelogram and similarly $BNDL$ is a parallelogram as well. But then $AM\parallel CK$ and $DL\parallel BN$, so $ABCD$ is a parallelogram. Therefore this case doesn't lead to any non-trivial solutions for $ABCD$.\newline\newline If WLOG $\frac{AW}{AK}=\frac{CY}{CM}=a$, then from (2) in $\textbf{Step 1}$ we have that: $$\frac{1}{5}S_{ABCD}=S_{\triangle AWZ}+S_{\triangle CXM}=a(S_{\triangle AZK}+S_{\triangle CXM})=\frac{2a}{5}(S_{\triangle ZXY}+S_{\triangle XWZ})=$$$$=\frac{2a}{5}S_{XYZW}=\frac{1}{5}S_{ABCD}\Longrightarrow a=\frac{2}{5}$$Thus $AW=\frac{2}{5}AK=WX$, but also $AN=NB$, so $DN\parallel BL$. It seems like we're so close to proving that $ABCD$ has to be a parallelogram, but we can make no further progress, due to the existence of counterexamples, which we presented in the next "chapter" of the solution. This was the thought process behind finding the counterexample. $\textbf{Counterexample:}$ Consider the coordinate plane and let $A=(0,0)$, $B=(1,1)$, $C=(3,-1)$, $D=(1,-2)$. The quadrilateral $ABCD$ is convex. Also $K=(2,0)$, $L=(2,-\frac{3}{2})$, $M=(\frac{1}{2},-1)$, $N=(\frac{1}{2},\frac{1}{2})$. Let $X=AK\cap BL$, $Y=BL\cap CM$, $Z=CM\cap DN$, $W=DN\cap AK$ be the common points of the segments $AK,BL,CM,DN$. Then $X,W\in AK\Longrightarrow X=(x_{0},0)$ and $W=(w_{0},0)$ for some $0<x_{0},w_{0}<2$ and $Y,Z\in CM\Longrightarrow Y=(y_{0},-1)$ and $Z=(z_{0},-1)$ for some $\frac{1}{2}<z_{0},y_{0}<3$. Now we will calculate $x_{0},y_{0},z_{0},w_{0}$. We'll do this by calculating the slope of the lines $DN$ and $BL$. Let's denote $\ell_{1}$ to be the line $DN$ and $BL$ to be the line $\ell_{2}$ for shortness. Firstly, let's calculate the slope of $\ell_{1}$ and let's call it $s_{1}$ (we'll denote by $(x_{t},y_{t})$ the point $T$ even if we know the exact coordinates of said point just so we write out the slope of line fomrula clearly): $$D\in\ell_{1}, N\in\ell_{1}\Longrightarrow s_{1}=\frac{y_{n}-y_{d}}{x_{n}-x_{d}}=\frac{\frac{1}{2}-(-2)}{\frac{1}{2}-1}=-5$$$$Z\in \ell_{1},D\in \ell_{1}\Longrightarrow -5=s_{1}=\frac{y_{z}-y_{d}}{x_{z}-x_{d}}=\frac{-1-(-2)}{z_{0}-1}=\frac{1}{z_{0}-1}\Longrightarrow z_{0}=\frac{4}{5}$$$$W\in \ell_{1},D\in \ell_{1}\Longrightarrow -5=s_{1}=\frac{y_{w}-y_{d}}{x_{w}-x_{d}}=\frac{0-(-2)}{w_{0}-1}=\frac{2}{w_{0}-1}\Longrightarrow w_{0}=\frac{3}{5}$$$$\Longrightarrow W=\left(\frac{3}{5},0\right),\quad Z=\left(\frac{4}{5},-1\right)$$$$B\in\ell_{2}, L\in\ell_{2}\Longrightarrow s_{2}=\frac{y_{b}-y_{l}}{x_{b}-x_{l}}=\frac{1-(-\frac{3}{2})}{1-2}=-\frac{5}{2}$$$$X\in \ell_{2},B\in \ell_{2}\Longrightarrow -\frac{5}{2}=s_{2}=\frac{y_{x}-y_{b}}{x_{x}-x_{b}}=\frac{0-1}{x_{0}-1}=\frac{-1}{x_{0}-1}\Longrightarrow x_{0}=\frac{7}{5}$$$$Y\in \ell_{2},B\in \ell_{2}\Longrightarrow -\frac{5}{2}=s_{2}=\frac{y_{y}-y_{b}}{x_{y}-x_{b}}=\frac{-1-1}{y_{0}-1}=\frac{-2}{y_{0}-1}\Longrightarrow y_{0}=\frac{9}{5}$$$$\Longrightarrow X=\left(\frac{7}{5},0\right),\quad Y=\left(\frac{9}{5},-1\right)$$Now we can calculate the ratios of the medial parts to the whole segments of $AK,BL,CM,DN$. For the segments $AK$ and $CM$ the calculations are easier, because all $y$ values of points on either of those lines are equal, but for $BL$ and $DN$ we'll need to use the Pythagorian theorem. Here are the claculations: $$AK=2, \quad XW=x_{x}-x_{w}=\frac{7}{5}-\frac{3}{5}=\frac{4}{5}\Longrightarrow \frac{XW}{AK}=\frac{2}{5}$$$$CM=x_{c}-x_{m}=3-\frac{1}{2}=\frac{5}{2},\quad YZ=x_{y}-x_{z}=\frac{9}{5}-\frac{4}{5}=1\Longrightarrow \frac{YZ}{CM}=\frac{2}{5}$$$$BL=\sqrt{(x_{b}-x_{l})^2+(y_{b}-y_{l})^2}=\sqrt{1+\frac{25}{4}}=\frac{\sqrt{29}}{2}$$$$XY=\sqrt{(x_{x}-x_{y})^2+(y_{x}-y_{y})^2}=\sqrt{\frac{4}{25}+1}=\frac{\sqrt{29}}{5}\Longrightarrow \frac{XY}{BL}=\frac{2}{5}$$$$DN=\sqrt{(x_{d}-x_{n})^2+(y_{d}-y_{n})^2}=\sqrt{\frac{1}{4}+\frac{25}{4}}=\frac{\sqrt{26}}{2}$$$$ZW=\sqrt{(x_{z}-x_{w})^2+(y_{z}-y_{w})^2}=\sqrt{\frac{1}{25}+1}=\frac{\sqrt{26}}{5}\Longrightarrow \frac{ZW}{DN}=\frac{2}{5}$$Therefore we've shown that $\frac{XW}{AK}=\frac{YZ}{CM}=\frac{XY}{BL}=\frac{ZW}{DN}=\frac{2}{5}$ and $ABCD$ isn't a parallelogram (the simplest proof of this is the fact that $AB=\sqrt{(x_{a}-x_{b})^2+(y_{a}-y_{b})^2}=\sqrt{2}$ and $CD=\sqrt{(x_{c}-x_{d})^2+(y_{c}-y_{d})^2}=\sqrt{5}$, so $AB\neq CD\Longrightarrow ABCD$ isn't a parallelogram), so the answer to the problem is negative, the conditions don't yield that $ABCD$ is a parallelogram, because we showed a counterexample.