Firstly, we prove that $BDLF$ is an isosceles trapezium (Fig. 44). As $\angle CEB = 90^\circ = \angle CFB$, points $B$, $C$, $E$ and $F$ lie on a circle with diameter $BC$. Hence $M$ is the center of this circle. Consequently, $MB = MF$, implying

Fig. 44

$DB = MB - MD = MF - ML = LF$. Hence $DL \parallel BF$ and the desired claim follows.

Consequently, points $B$, $D$, $L$ and $F$ also lie on a circle (Fig. 45). Since $\angle BDH = 90^\circ = \angle HFB$, we know that $BH$ is a diameter of this circle. Thus $H$ lies on this circle, too. Now
$$\begin{aligned} \angle MLK &= \frac{\angle 180^\circ - \angle KML}{2} = \frac{\angle LME}{2} = \frac{\angle FME}{2} \\ &= \angle FBE = \angle FBH = \angle FLH. \end{aligned}$$

But $\angle MLK = \angle FLH$ implies that points $K$, $L$, and $H$ lie on a line.

As in Solution 1, we prove that $BDLF$ is an isosceles trapezium whose circumcircle passes through $H$. By interchanging the roles of $B$ and $C$, the roles of $E$ and $F$, and the roles of $K$ and $L$, we analogously obtain that $CKDE$ is an isosceles trapezium whose circumcircle passes through $H$ (Fig. 46). Now $\angle LHB = \angle LFB = \angle FBD = \angle ABC$, but, on the other hand,
$$\angle KHB = 180^\circ - \angle EHK = \angle KCE = \angle CED.$$
As $\angle BDA = 90^\circ = \angle BEA$, the quadrilateral $ABDE$ is cyclic, implying that $\angle CED = \angle ABC$. Consequently, $\angle LHB = \angle KHB$, implying that points $K$, $L$ and $H$ lie on a line.

As in Solution 1, we prove that points $B$, $D$, $L$, $F$ and $H$ lie on a circle with diameter $BH$. By interchanging the roles of $B$ and $C$, the roles of $E$ and $F$, and the roles of $K$ and $L$, we analogously obtain that points $C$, $D$, $K$, $E$ and $H$ lie on a circle with diameter $CH$. As $HL \perp BL$ and $HK \perp CK$ by Thales' theorem (Fig. 47), it suffices to prove that $BL \parallel CK$. To this end, note that
$$\begin{aligned} \angle LBM &= \angle LBD = \angle LFD = \angle MFD, \\ \angle KCM &= \angle KCD = \angle KED = \angle MED. \end{aligned}$$

Fig. 46

Fig. 47

It is known that the midpoints of sides of a triangle and the feet of altitudes of the triangle lie on a circle (so-called nine-point circle). Hence $M$, $D$, $E$ and $F$ are concyclic, implying that $\angle MFD = \angle MED$. Consequently, $\angle LBM = \angle KCM$, proving the desired result.