The expression under investigation can be written as

$$\begin{gathered} K=\left(y_{1}+y_{4}\right)-\left(y_{2}+y_{3}\right)=\left(y_{4}-y_{3}\right)-\left(y_{2}-y_{1}\right)=c\left(x_{4}^{2}-x_{3}^{2}\right)- \\ \quad-c\left(x_{2}^{2}-x_{1}^{2}\right)=c\left(x_{4}-x_{3}\right)\left(x_{4}+x_{3}\right)-c\left(x_{2}-x_{1}\right)\left(x_{2}+x_{1}\right)= \\ =\operatorname{ch}(2 b+2 d+h)-\operatorname{ch}(2 b-2 d-h)=\operatorname{ch}(4 d+2 h)=2 \operatorname{ch}(2 d+h) \end{gathered}$$

The final form indeed does not contain $b$, indicating independence from $b$.

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It is visible that the points $A_{1}, A_{4}$ and $A_{2}, A_{3}$ on the $X$-axis with abscissae $x_{1}$ and $x_{4}$, and $x_{2}$ and $x_{3}$, respectively, are symmetric pairs with respect to the abscissa $b$. Accordingly, the midpoints of the chords $F_{14}$ and $F_{23}$ connecting the corresponding points $P_{1}$ and $P_{4}$, and $P_{2}$ and $P_{3}$ on the graph of the function $y=c x^{2}$, are also on the abscissa $b$, one above the other. The

$$\frac{K}{2}=\frac{y_{1}+y_{4}}{2}-\frac{y_{2}+y_{3}}{2}=\operatorname{ch}(2 d+h)$$

value precisely gives the height difference, and thus the distance, between the two midpoints. The two chords are parallel because

$$\frac{y_{4}-y_{1}}{x_{4}-x_{1}}=c\left(x_{4}+x_{1}\right)=2 b c=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}$$

Accordingly, translating the point system $A_{1}, A_{2}, A_{3}, A_{4}$ as a rigid body along the $X$-axis changes the slopes of the chords $P_{1} P_{4}$ and $P_{2} P_{3}$, but the distance between their midpoints remains constant.

Another transformation yields

$$\begin{gathered} \frac{K}{h}=\frac{y_{4}-y_{3}}{h}-\frac{y_{2}-y_{1}}{h}=\frac{y_{4}-y_{3}}{x_{4}-x_{3}}-\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=2 c(2 d+h)= \\ =2 c\left(x_{3}-x_{1}\right)=2 c\left(x_{4}-x_{2}\right) \end{gathered}$$

From this, we can read that the change in the slope of the chord $P_{1} P_{2}$ when the rigid point pair $A_{1}, A_{2}$ is translated to $A_{3}, A_{4}$ is proportional to the translation $x_{3}-x_{1}$. Naturally, the same is given by

$$\frac{K}{2 d+h}=\frac{y_{4}-y_{2}}{x_{4}-x_{2}}-\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=2 c h=2 c\left(x_{2}-x_{1}\right)$$

for the translation of the point pair $A_{1}, A_{3}$ to $A_{2}, A_{4}$.

For $d=0$, we have $x_{2}=x_{3}=b=\left(x_{1}+x_{4}\right) / 2, y_{2}=y_{3}$, and

$$\frac{K}{2}=\frac{y_{1}+y_{4}}{2}-y_{2}=c h^{2}$$

Accordingly, if the midpoint of the segment $A_{1} A_{4}$ is $A_{2}$, then the distance from the midpoint of the chord $P_{1} P_{4}$ to $P_{2}$ does not change with the translation of the rigid point system $A_{1}, A_{2}, A_{4}$. The distance is proportional to the square of the segment $A_{1} A_{2}$.

Alternatively, expressing each $y$ ordinate in terms of the corresponding abscissa and dividing by $c$,

$$\left(\frac{K}{2 c}=\right) \frac{x_{1}^{2}+x_{4}^{2}}{2}-x_{2}^{2}=h^{2}, \quad \frac{x_{1}^{2}+x_{4}^{2}}{2}=\left(\frac{x_{1}+x_{4}}{2}\right)^{2}+\left(\frac{x_{4}+x_{1}}{2}\right)^{2}$$

the arithmetic mean of the squares of two numbers is greater than the square of their mean by the square of half their difference.