Given the trapezoid $ABCD\ :\ A(0,0),B(b,0),C(b-c,d),D(c,d)$. Points $E(\frac{kb+b-c}{1+k},\frac{d}{1+k})$ and $F(\frac{kc}{1+k},\frac{kd}{1+k})$ with $k<1$. Points $L(\lambda,0)$ and $L(\frac{kc+kb+b-c}{1+k}-\lambda,d)$. Slope of the line $KE\ :\ m_{KE}=\frac{d}{kb+b-c-\lambda(1+k)}=m_{FL}$, the slope of the line $FL$, so $KE // FL$. Slope of the line $KF\ :\ m_{KF}=\frac{kd}{kc-\lambda(1+k)}=m_{LE}$, the slope of the line $LE$, so $KF // LE$.

Here is a nicer solution: Let $X$ be the midpoint of $EF$, hence also the midpoint of $MN$. We need to prove that $X$ is the midpoint of $KL$ i.e. that $MKNL$ is a parallelogram. Clearly $MK \parallel NL$, so it suffices to prove $MK=NL$. But this follows by an easy trick: Reflect over the perpendicular line to $AB$ through $F$ and let $M'$ and $K'$ be the images of $M,K$. Then clearly $M'FAK'$ is congruent to $NECL$ and hence $NL=M'K'=MK$ as desired.

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solution later

SEE THE FIGURE:

Attachments:

1.claim: FE is middle line. prove:let AD=BC=a , BE=DF=x , FA=BE from Tales so a-x=x => a=2x => AF=FD=BE=EC so FE is middle lie. 2.claim: triangle XKM and triangle XLN are same. (X be point intersection of KL and FE) prove: $\angle$ AKM = $\angle$ FMK = 90 because AB || FE similar $\angle$FNL=90 so MK || NL. and MK = NL from Tales and $\angle$ MKX =$\angle$ NLX because MK || NL. KX=XL from Tales => => so we can say triangle XKM and triangle XLN are same.And MX=XN=>FX=XE (1) KX=XL=(2) (1)=>(2)=>> FKLE is parallelogram.

We are to show that MK=NL. Draw a line parallel to AD from E intersecting CD at K'. Then, we will get EK'=EC=AF, we also have <NEK'=<AFM and NE=FM; so, tr. NEK' and tr. AFM are congruent ,NK'=AM, <NK'E=<FAM. We can easily see that quad. AKMF and quad. NEK'L have equal angles , <A=<EK'L now we can say <KAM=<NK'L, we have <AKM=NLK'=90° so, tr. AKM is congruent to tr. NK'L and hence MK=NL.