Cesko-Slovacko-Poljsko

Denote by $S$ the midpoint of $AB$, by $M$ the midpoint of $KL$, and by $P$ and $Q$ the feet of the altitudes from $A$ and $B$ in the triangle $ABC$. Obviously, $P$ and $Q$ are the midpoints of $AK$ and $BL$ respectively (fig. 3). Therefore $QS$ is the mid-segment in the triangle $LAB$ and $MP$ is the mid-segment in the triangle $LAK$. We have $$QS=\frac{1}{2}LA,MP=\frac{1}{2}LA\text{and}QS\parallel LA\parallel MP.$$ thus $SPMQ$ is a parallelogram (this is true even in the case of "degenerate" triangles $LAB$ or $LAK$). Fig. 3 The points $P$ and $Q$ lie on the Thales circle over $AB$, so $SP=SQ=\frac{1}{2}AB$. This implies the parallelogram $SPMQ$ is a rhombus and the length of its side is independent of the location of $C$. To prove that also the length of its diagonal $SM$ is independent of $C$, it suffices to show that the size of the angle spanned by its sides $SP$ and $SQ$ is constant when moving $C$ along $k$ (then all the rhombuses $SPMQ$ and also their diagonals $SM$ are congruent). Fig. 4a Fig. 4b If the angle $\alpha$ in the triangle $ABC$ is acute, the point $Q$ lies inside of $AC$ (the angle $\gamma$ is always acute by the statement of the problem) and the angle $PSQ$ is central to the angle $PAQ$, which is inscribed over the chord $PQ$ of the Thales circle over $AB$ (fig. 4a). Hence $$\angle PSQ=2\angle PAQ=2(90^{\circ }-\gamma )=180^{\circ }-2\gamma .$$

The same formula we get when $\alpha$ is non-acute, since in this case $\beta$ is acute and we can use the inscribed angle $PBQ$ instead of $PAQ$ (fig. 4b): $$\angle PSQ=2\angle PBQ=2(90^{\circ }-\gamma )=180^{\circ }-2\gamma .$$ As the size of $\gamma$ does not change when moving $C$ along $k$ (it is an inscribed angle over the fixed chord $AB$), the size of the angle $PSQ$ also does not change.

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Alternative solution.

Denote by $\alpha ,\beta ,\gamma$ the angles of the triangle $ABC$. We only consider the case $\alpha <90^{\circ }$; the second case ($\beta <90^{\circ }$) is similar. Let $S,M,U,V$ be the midpoints of the segments $AB,KL,AL,BK$ respectively. Let $H$ be the common point of the lines $AK$ and $BL$ (fig. 5a, b). Note that the quadrilateral $USVM$ is a parallelogram ($SU\parallel BL\parallel MV,SV\parallel AK\parallel MU$) and $$SV=AB\sin \beta ,MV=SU=AB\sin \alpha ,\angle SVM=\angle AIL=\angle ACB$$ By the law of sines applied to the triangle $ABC$, $$\frac{AC}{SV}=\frac{1}{AB}\cdot \frac{AC}{\sin \beta }=\frac{1}{AB}\cdot \frac{BC}{\sin \alpha }=\frac{BC}{MV}$$ so the triangles $SVM$ and $ACB$ are proportional (two sides proportional and the same angle between them). Then again by the law of sines. $$SM=AB\cdot \frac{SV}{AC}=\frac{A{B}^2\sin \beta }{AC}=\frac{A{B}^2\sin \gamma }{AB}=AB\sin \gamma$$ which obviously does not depend of the location of the point $C$. Fig. 5a Fig. 5b