Estonian Mathematical Olympiad

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 $$\begin{array}{rl}\angle MLK& =\frac{\angle 180^{\circ }-\angle KML}{2}=\frac{\angle LME}{2}=\frac{\angle FME}{2}\\ & =\angle FBE=\angle FBH=\angle FLH.\end{array}$$

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

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Alternative solution.

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.

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Alternative solution.

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 $$\begin{array}{rl}\angle LBM& =\angle LBD=\angle LFD=\angle MFD,\\ \angle KCM& =\angle KCD=\angle KED=\angle MED.\end{array}$$

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.