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
$$QS = \frac{1}{2} LA, \quad MP = \frac{1}{2} LA \quad \text{and} \quad QS \parallel LA \parallel MP.$$
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.

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
$$SV = AB \sin \beta, \quad MV = SU = AB \sin \alpha, \quad \angle SVM = \angle AIL = \angle ACB$$
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.
$$SM = AB \cdot \frac{SV}{AC} = \frac{AB^2 \sin \beta}{AC} = \frac{AB^2 \sin \gamma}{AB} = AB \sin \gamma$$
which obviously does not depend of the location of the point $C$.

Fig. 5a

Fig. 5b