Let $A(a, a^2)$, $B(b, b^2)$, $C(c, c^2)$, $D(d, d^2)$, $K(k, k^2)$, $L(l, l^2)$, $M(m, m^2)$, $N(n, n^2)$. Without loss of generality we can assume that the points are arranged as shown in the figure (all other cases are similar).

It is easy to write the equations of the straight lines $AB$, $CD$, $KL$, $AD$, $BC$
$$AB: y = (a+b)x - ab, \quad CD: y = (c+d)x - cd, \quad KL: y = (k+l)x - kl,$$
$$AD: y = (a+d)x - ad, \quad BC: y = (b+c)x - bc.$$
Since the lines $AB$, $CD$, $KL$ are parallel, their slopes are equal, so
$$a+b=c+d=k+l=\lambda, \quad (1)$$
where $\lambda$ is some real number.

The point $M$ is the intersection point of the lines $AD$ and $KL$, so its abscissae $m$ satisfies the equation
$$(k+l)m - kl = (a+d)m - ad \Leftrightarrow \\ \Leftrightarrow (k+l-a-d)m = kl-ad.$$
Taking into account (1), we obtain
$$(c-a)m = kl-ad. \quad (2)$$

Similarly, the abscissae $n$ of the point $N$ ($N$ is the intersection point of the lines $BC$ and $KL$) satisfies the equation
$$(k+l)n - kl = (b+c)n - bc \Leftrightarrow (k+l-b-c)n = kl-bc$$
Taking into account (1), we obtain
$$(a-c)n = kl - bc. \quad (3)$$
Subtracting (3) from (2), we get
$$(c-a)(m+n) = bc-ad = [b \stackrel{(1)}{=} \lambda-a, d \stackrel{(1)}{=} \lambda-c] = (\lambda-a)c-a(\lambda-c) = \lambda(c-a).$$
Since $c - a \neq 0$, we have $m + n = \lambda \stackrel{(1)}{=} k + l$, i.e., $m + n = k + l$. This equality can be written as $\frac{m+n}{2} = \frac{k+l}{2}$. It follows that the midpoints of the segments $MN$ and $KL$ coincide. Since $MN$ and $KL$ lie on the same straight line, we obtain $KM = NL$ as required.