Find the geometric locus of midpoints of segments connecting a given point with points lying on a given plane. Exercise

Find the geometric locus of midpoints of segments connecting a given point with points lying on a given plane.


Attempt
Given


*

*Point $P$

*Plane $M$


Construction
Mark an arbitrary point $Q$ on $M$.
Draw line segment $l$; $l$=$PQ$.
Bisect $l$ at some resultant point $R$.
Draw plane $N$; $N || M$, $R$ lies on $N$.
Answer
$N$ is the geometric locus of midpoints of segments connecting a given point with points lying on a given plane.

Question

I can give you the answer, as well as the construction, however I don't know how to prove that my solution is legit.


Postscript


*

*This exercise is found in Kiselev's Geometry; Book II: Stereometry (English adaptation).


*

*It is Exercise 17, in Chapter 1: Lines and planes, Section 2: Parallel lines and planes.


 A: Your transformation is an homothetic transformation $h$ whose center is the given point $S$ and ratio $\lambda=\frac{1}{2}$.
The image of the plane by $h$ is another plane. More precisely the image is the plane passing through the images of three independent points lying on the initial plane.
A: I complement the (very accurate) solution of @
mathcounterexamples.net by an analytical proof:
If the plane $(P)$ has equation 
$$\tag{1}ux+vy+wz=h$$
and the point, say $P_0$, has coordinates $(x_0,y_0,z_0)$, the locus of midpoints is the set of points having coordinates $$\begin{cases}X&=&\frac{x+x_0}{2}\\Y&=&\frac{y+y_0}{2}\\Z&=&\frac{z+z_0}{2}\end{cases} \ \ \ \Leftrightarrow \ \ \ \begin{cases}x&=&2X-x_0\\y&=&2Y-y_0\\z&=&2Z-z_0\end{cases}$$
Plugging these expressions into (1), we get:
$$u(2X-x_0)+v(2Y-y_0)+w(2Z-z_0)=h$$
which induces the following constraint on the set of points $(X,Y,Z)$:
$$\tag{2} uX+vY+wZ=\dfrac{1}{2}(ux_0+vy_0+wz_0)$$
The geometrical interpretation of (2) is that the set of points $(X,Y,Z)$ belongs to a certain plane $(P')$ parallel to the initial plane $(P)$ (they share the same normal vector $(u,v,w)$).
Moreover, as we have proceeded by equivalent conditions, the solution is the whole plane.
