Solution:

Let $O$ be the intersection of the diagonals. Let $P_1$ be on $AB$ such that $PP_1 \parallel AD$, and let $Q_1$ be on $CD$ such that $QQ_1 \parallel AD$. Let $\sigma$ be the central symmetry with center $O$. Let $P' = \sigma(P)$, $Q' = \sigma(Q)$, $P_1' = \sigma(P_1)$.

Let $M_1 = AQ_1 \cap DP_1$, $N_1 = BQ_1 \cap CP_1$, $N' = AQ' \cap DP'$, and $M' = BQ' \cap CP'$. Then: $M' = \sigma(M)$, $N' = \sigma(N)$, $M_1' = \sigma(M_1)$, and $N_1' = \sigma(N_1)$.

Since $AP$ and $DP_1$ are the diagonals of the parallelogram $AP_1PD$, $CP_1$ and $BP$ are the diagonals of the parallelogram $P_1BCP$, and $AQ_1$ and $DO$ are the diagonals of the parallelogram $AQQ_1D$, it follows that the points $U, V, W$ (figure 2) are collinear and they lie on the line passing through the midpoints $R$ of $AD$ and $Z$ of $BC$. The diagonals $AM$ and $DM_1$ of the quadrilateral $AM_1MD$ intersect at $U$ and the diagonals $AM_1$ and $DM$ intersect at $W$. Since the midpoint of $AD$ is on the line $UW$, it follows that the quadrilateral $AM_1MD$ is a trapezoid. Hence, $MM_1$ is parallel to $AD$ and the midpoint $S$ of $MM_1$ lies on the line $UW$ (figure 2).

Figure 11

Similarly $M'M_1'$ is parallel to $AD$ and its midpoint lies on $UW$. So $M_1M'M_1'M$ is a parallelogram whose diagonals intersect at $O$.

Similarly, $N_1'N N_1N'$ is a parallelogram whose diagonals intersect at $O$.

All these imply that $M, N, M', N'$ and $O$ are collinear, i.e. $O$ lies on the line $KL$. This implies that $K = \sigma(L)$, and since $D = \sigma(B)$, the conclusion follows.

Figure 12

Solution:

Let the line $\varepsilon$ through the points $M, N$ intersect the lines $DC, AB$ at points $T_1, T_2$ respectively. Applying Menelaus' Theorem to the triangles $DQC$, $APB$ with intersecting line $\varepsilon$ in both cases we get:

$$\frac{MD}{MQ} \cdot \frac{NQ}{NC} \cdot \frac{T_1C}{T_1D} = 1 \quad \text{and} \quad \frac{MP}{MA} \cdot \frac{NB}{NP} \cdot \frac{T_2A}{T_2B} = 1$$

But it is true that $\frac{MD}{MQ} = \frac{MP}{MA}$ and $\frac{NQ}{NC} = \frac{NB}{NP}$.

It follows that $\frac{T_1C}{T_1D} = \frac{T_2A}{T_2B}$, i.e. $\frac{T_1D + DC}{T_1D} = \frac{T_2B + BA}{T_2B}$, hence $T_1D = T_2B$ (since $DC = BA$).

Then of course by the similarity of the triangles $T_1DK$ and $T_2BL$ we get the desired equality $DK = BL$.

Figure 13