$\rightarrow (KD,LC)=J$. $\rightarrow \frac{FA}{FD}=\frac{AE}{EC} \Rightarrow \frac{FC}{FD}=\frac{FC}{FA}.\frac{AE}{EC}=\frac{\sin(\widehat{DFE})}{\sin(\widehat{CFE})}$. $\Rightarrow FE$ is $F$-median of $\triangle{FDC}$. $\rightarrow \frac{JD}{DK} = \frac{CE}{EA} = \frac{BE}{ED} = \frac{LC}{CJ} \Rightarrow {KL} \parallel {DC}$. $\rightarrow \frac{AE}{DJ} = \frac{AE}{EC} = \frac{BE}{ED} = \frac{BE}{CJ}.\frac{AB}{CD} = \frac{FA}{FD} = \frac{FB}{FC}$. (a) $\rightarrow \widehat{FAE} = \widehat{FDJ}$ and $\widehat{FBE} = \widehat{FCJ}$. (b) By (a) and (b) we get that polygon $FAEB$ is dilated with $F$ as the center of dilation resulting in the image $FDJC \Rightarrow F,E,J$are collinear. $\Rightarrow JE$ is the median of $\triangle{DJC} \Rightarrow JE$ is the median of $\triangle{KJL} \blacksquare$.

\[\begin{gathered} If:\mathop {}\limits^{} I = AK \cap BL\mathop {}\limits^{} ,\mathop {}\limits^{} H = KD \cap LC\mathop {}\limits^{} ,\mathop {}\limits^{} O = EI \cap AB\mathop {}\limits^{} ,\mathop {}\limits^{} N = EH \cap DC\mathop {}\limits^{} ,\mathop {}\limits^{} M = EH \cap KL\mathop {}\nolimits_{} \mathop {}\limits_{} then\mathop {}\limits_{}^{} : \hfill \\ AIBE\mathop {}\nolimits^{} ,\mathop {}\nolimits^{} EDCH\mathop {}\nolimits^{} ,\mathop {}\limits_{} IKHL\mathop {}\nolimits^{} \mathop {}\nolimits^{} are\mathop {}\nolimits^{} similar\mathop {}\nolimits^{} paralle\log rams\mathop {}\nolimits^{} ,\mathop {}\nolimits^{} so\mathop {}\nolimits^{} ,\mathop {}\nolimits^{} I,E,H\mathop {}\nolimits^{} ,\mathop {}\nolimits^{} are\mathop {}\nolimits^{} collinear \hfill \\ and\mathop {}\nolimits^{} \mathop {}\nolimits^{} therefore\mathop {}\nolimits^{} \mathop {}\nolimits^{} IH\mathop {}\nolimits^{} passes\mathop {}\limits^{} \mathop {}\nolimits^{} from\mathop {}\nolimits^{} \mathop {}\nolimits^{} F. \hfill \\ \end{gathered} \]

Attachments:

Dear Mathlinkers, 1. I consider the figure above 2. according to a special case of the Pappus theorem, JL//AB 3. according two time the theorem of complete quadrilateral(trapezoid), we are done. Sincerely Jean-Louis

Let $A=0$, $B=1$, $C=c$, $D=c+t$, where $c \in \mathbb{C}$, $t \in \mathbb{R}$. Because $A,D,F$ are collinear, $F=(c+t)p$, $p \in \mathbb{R}$. Also $\frac{(c+t)p-1}{(\overline c + t)p-1}=\frac{c-1}{\overline c - 1} \Rightarrow F = \frac{c+t}{t+1}$. Similarly $E=cq$, $q \in \mathbb{R}$ and $c+t$, $cq$, $1$ are collinear, so $\frac{c+t-1}{\overline c + t-1} = \frac{cq-1}{\overline cq-1} \Rightarrow E = \frac{c}{1-t}$. $AEDK$ and $BECL$ are parallelograms, so $E+K=D \Rightarrow K = c+t-\frac{c}{1-t}$ and $(E-B)+(L-B)=C-B \Rightarrow L = c-\frac{c}{1-t}+1$. The midpoint of the segment $KL$ is $M=c-\frac{c}{1-t}+\frac{1}{2}+\frac{1}{2}t$. Now we can prove that $\frac{M-E}{\overline M - \overline E}=\frac{F-E}{\overline F - \overline E}$, which completes the proof.

We are going to use Barycentric Coordinates. Let $F$ be the intersection of line $AD$ and $BC$. Also, let $FDC$ be the reference triangle. So, $F(1, 0, 0); D(0, 1, 0); C(0, 0, 1)$. Let $B(u, 0, 1-u)$. Since $AB || CD$, $A(u, 1-u, 0)$. $E$ is the intersection of line $CA$ and $DB$, and thus has $x:y$ and $x:z$ coordinate ratio of $(u:1-u)$. Thus $E(u:1-u:1-u)$ or $E(\frac{u}{2-u}, \frac{1-u}{2-u}, \frac{1-u}{2-u})$. Since $AEDK$ and $BECL$ are parallelograms, we know that $K = A + D - E$, so $K(u - \frac{u}{2-u}, 2-u-\frac{1-u}{2-u}, -\frac{1-u}{2-u})$. Similarly, $L(u - \frac{u}{2-u}, -\frac{1-u}{2-u}, 2-u-\frac{1-u}{2-u})$. Thus $M = \frac{(K + L)}{2}$ and has coordinate of $M(u - \frac{u}{2-u}, \frac{2-u}{2} - \frac{1-u}{2-u}, \frac{2-u}{2} - \frac{1-u}{2-u})$. Since both $E$ and $M$ has $y:z$ coordinate ratio of $1:1$, then $F, E, M$ colinear.

Denote $F=(0,0), A = (2,2a), B = (2,2b), C = (2k,2kb), D = (2k,2ka)$. We first find the coordinate of $E$. Line $BD$ has the equation $y = \frac{ka-b}{k-1} (x-2) + 2b$ and line $AC$ has the equation $y = \frac{kb-a}{k-1} (x-2) + 2a$. Solving gives $E = \left( \frac{4k}{k+1}, \frac{2k(a+b)}{k+1} \right)$. Because $AEPK$ is a parallelogram, we can get $K = \left( \frac{2k^2+2}{k+1} , \frac{2ak^2+2ak-2bk+2a}{k+1} \right)$ and $L = \left( \frac{2k^2+2}{k+1} , \frac{2bk^2+2bk-2ak+2b}{k+1} \right)$. Then, $\left( \frac{2k^2+2}{k+1} , \frac{(k^2+1)(a+b)}{k+1}\right)$. Notice that gradient of $EF$ and $EM$ is both $\frac{a+b}{2}$, so we are done