In the solution, we will use the concept of the power of a point with respect to a circle, as well as the power line of two circles, and their fundamental properties. Definitions and statements related to this can be found, for example, in the 923rd and 931st problems of the Collection of Geometric Problems II.

First, we prove an auxiliary theorem:

(*) Let $X$ and $Y$ be arbitrary points on the line segments $AC$ and $BC$ of triangle $ABC$. Then the power line of the circles with diameters $AY$ and $BX$ contains the orthocenter $M$ of triangle $ABC$.

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Figure 1

Let the feet of the altitudes from $A$ and $B$ to the opposite sides of triangle $ABC$ be $A_1$ and $B_1$, respectively, and let the circles with diameters $AY$ and $BX$ be denoted by $k_A$ and $k_B$ (Figure 1). Then $\angle AA_1Y = \angle BB_1X = 90^\circ$, so point $A_1$ lies on circle $k_A$, and point $B_1$ lies on circle $k_B$; furthermore, points $A_1$ and $B_1$ lie on the circle $k$ with diameter $AB$. Therefore, the power of point $M$ with respect to circles $k_A$ and $k_B$ is $MA \cdot MA_1$ and $MB \cdot MB_1$, respectively, which are equal because both are equal to the power of point $M$ with respect to circle $k$. Thus, point $M$ lies on the power line of the two circles.

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Now let's turn to the solution of our original problem. Let the 6 intersection points of the 4 lines be denoted as $P, Q, R, S, T, U$ as shown in Figure 2. Choose 2 pairs of opposite points from the 6 intersection points - that is, points that do not lie on the same line - in our diagram, $P, U$ and $S, Q$. We claim that the power line of the circles with diameters $PU$ and $SQ$ contains all four orthocenters. This follows from our auxiliary theorem, if we apply it four times with the following choices:

$$\begin{aligned} & \text{1. 2. 3. 4.} \\ & A: P P S Q \\ & B: S Q U U \\ & C: R T T R \\ & X: Q S P P \\ & Y: U U Q S \end{aligned}$$

With this, we have proven the statement of our problem.