41. It is well known that the points $K, L, M$, symmetric to $H$ with respect to $B C, C A, A B$ respectively, lie on the circumcircle $k$ of the triangle $A B C$. For $K$, this follows from an elementary calculation of angles of triangles $H B C$ and noting that $\measuredangle K B C=\measuredangle H B C=\measuredangle K A C$. For other points the proof is analogous. Since the lines $l_{a}, l_{b}$ pass through $K$ and $L$ and $l_{b}$ is obtained from $l_{a}$ by rotation about $C$ for an angle $2 \gamma=\angle L C K$, it follows that the intersection point $P$ of $l_{a}$ and $l_{b}$ is at the circumcircle of $K L C$, that is, $k$. Similarly, $l_{b}$ and $l_{c}$ meet at a point on $k$; hence they must pass through the same point $P$. !