Consider a right-angled triangle as shown below, where vertex $O$ is the right angle. If we consider the midpoint of the two legs and draw a line passing through it, we have proved that the area of triangle $OCD$ is one-fourth of the area of triangle $OAB$.
Similarly, for a pyramid with a triangular base and vertex $O$ as shown in the figure below, if we pass a plane through the midpoint of all the edges, the volume of pyramid $ODEF$ is one-eighth of the volume of pyramid $OABC$.
We want to calculate the ratio of volumes in higher dimensions. Perhaps the expression I use may not be mathematically accurate, but I hope to convey my intention correctly. We want to consider a right-angled triangle in $D$-dimensional space, pass a $D-1$ dimensional triangle through the legs, and calculate the ratio of the newly created $D$-dimensional triangle to the original one. Considering that in two-dimensional space the ratio is $\dfrac{1}{4}$, and in three-dimensional space, it is $\dfrac{1}{8}$, we conjecture that this rule for a shape in $D$ dimensions is equal to $\dfrac{1}{2^D}$. To prove this for 3-dimensional, we proceeded as follows:
Our solution
Consider the following image: The points of intersection of the pyramid with the coordinate axes are named $A, B$, and $C$. The coordinates of these points are as follows:
Point $A: (2x_0, 0, 0)$
Point $B: (0, 2y_0, 0)$
Point $C: (0, 0, 2z_0)$
Similarly, the coordinates of the midpoint points are named $A', B'$, and $C'$:
Point $A': (x_0, 0, 0)$
Point $B': (0, y_0, 0)$
Point $C': (0, 0, z_0)$
Now let us denote the volume $\chi$ as $V(\chi)$, then
$$ \dfrac{V(OA'B'C')}{V(AOBC)} = \dfrac{1/6 \text{ of the volume of the parallelepiped } A'B'C'}{1/6 \text{ of the volume of the parallelepiped } ABC} $$
which in turn can be calculated by the following determinant relation:
$$ \dfrac{1/6\det(\begin{bmatrix} x_0 & 0 & 0 \\ 0 & y_0 & 0 \\ 0 & 0 & z_0 \end{bmatrix})}{1/6\det(\begin{bmatrix} 2x_0 & 0 & 0 \\ 0 & 2y_0 & 0 \\ 0 & 0 & 2z_0 \end{bmatrix})} = \dfrac{1}{2^3} $$
Now, in the case of a 4-dimensional space (although it cannot be visualized), we consider four coordinate axes X, Y, Z, and W. By extending the concept from the 3-dimensional case, we obtained the ratio of volumes as follows:
$$ \dfrac{1/6\det(\begin{bmatrix} x_0 & 0 & 0 & 0 \\ 0 & y_0 & 0& 0 \\ 0 & 0 & z_0&0 \\ 0 & 0 &0 & w_0 \end{bmatrix})}{1/6\det(\begin{bmatrix} 2x_0 & 0 & 0 & 0 \\ 0 & 2y_0 & 0& 0 \\ 0 & 0 & 2z_0&0 \\ 0 & 0 &0 & 2w_0 \end{bmatrix})} = \dfrac{1}{2^4} $$
The question is: Is our method of proof and reasoning correct, and is the ratio of the two volumes in the D-dimensional space equal to $\dfrac{1}{2^D}$?
We have to understand what is volume $V(X)$ of a subset $X$ of $n$-dimensional space $\mathbb R^n$ for natural $n$. I think usually $V(X)$ is understood as ($n$-dimensional) Jordan measure $m(X)$ of $X$. The definition of Jordan measure is rather technical and requires rather advanced mathematical apparatus. But the required conclusion holds. Namely, if $n$ is any natural number, $\lambda>0$ is any real number, $X$ is any subset of $\mathbb R^n$, and $m(X)<\infty$ exists then $m(\lambda X)$ of the homothetic copy $\lambda X$ of $X$ exists and $m(\lambda X)=\lambda^n\mu(X)$.