Angles of triangle $\triangle XYZ$ do not depend on the position of point $P$ (proof needed)

Let $ABCD$ be a fixed convex quadrilateral and $P$ be an arbitrary point. Let $S,T,U,V,K,L$ be the projections of $P$ on $AB,CD,AD,BC,AC,BD$ respectively. Let $X,Y,Z$ be the midpoints of $ST,UV,KL$. Is it true that the angles of triangle $\triangle XYZ$ do not depend on the position of $P$?

I drew a figure on my computer and it seems that the angles of triangle $\triangle XYZ$ do not change with the position of point P:

(my original research)

Here is the proof using Mathematica and simple analytical geometry. In theory, you can do all this by hand, there is nothing special in it. But in real life, you need a computer.

Math involved in the proof is elementary and I did not have to use trigonometry until the very last step.

WLOG, we can assume that $A(0,0)$, $B(0, 1)$, $C(x_C,y_C)$, $D(x_D,y_D)$, $P(x_P,y_P)$:

a = {0, 0}; b = {0, 1}; c = {xc, yc}; d = {xd, yd}; p = {xp, yp};

We need just two functions. The first one calculates dot product of two vectors:

dotProduct[a_, b_] := Sum[a[[i]] b[[i]], {i, 1, Length[a]}];

The second one is used to find the coordinates of projection of point P to line AB:

projection[a_, b_, p_] := Module[
   {c, xc, yc, s, k},
   c = {xc, yc};
   s = Solve[{c - a == k (b - a), 
      dotProduct[b - a, c - p] == 0}, {xc, yc, k}];
    Return[c /. s[[1]]]; 
   ];

That's all that we need. The rest is simple and completely straightforward. We are going to "read" the picture and calculate coordinates of all the points.

First, let's calculate coordinates of points $S,T,U,V,K,L:$

s = projection[a, b, p];
t = projection[c, d, p];
u = projection[a, d, p];
v = projection[b, c, p];
k = projection[a, c, p];
l = projection[b, d, p];

Coordinates of points $X,Y,Z$ are:

x = (s + t)/2;
y = (u + v)/2;
z = (k + l)/2;

We need two vectors: $\vec{XY},\vec{XZ}$:

xy = y - x;
xz = z - x;

And, finally, we are ready to calculate the angle $YXZ$ from the formula:

$\cos^2\angle YXZ=\frac{(\vec{XY} \cdot \vec{XZ})^2}{(\vec{XY} \cdot \vec{XY})(\vec{XZ} \cdot \vec{XZ})}$

...or in Mathematica:

cosXsquared = Together[dotProduct[xy, xz]^2/(dotProd[xy, xy] dotProduct[xz, xz])]

And the final result is:

$$\cos^2\angle YXZ=\\\frac{\left({x_C}^2 {x_D}^2+{x_C}^2 {y_D}^2-{x_C}^2 {y_D}+{x_C} {x_D}+{x_D}^2 {y_C}^2-{x_D}^2 {y_C}+{y_C}^2 {y_D}^2-{y_C}^2 {y_D}-{y_C} {y_D}^2+{y_C} {y_D}\right)^2}{\left({x_C}^2+{y_C}^2\right) \left({x_C}^2+{y_C}^2-2 {y_C}+1\right) \left({x_D}^2+{y_D}^2\right) \left({x_D}^2+{y_D}^2-2 {y_D}+1\right)}$$

Depsite the fact that $x_P,y_P$ can be easily observed in all intermediate results, the coordinates of point $P$ have magically disappeared from the final one.

In the same way you can prove that the other two angles of triangle $\triangle XYZ$ are independent from the location of point $P$.

Hint: One way to solution is with a use of a following lemma:

Lemma: Let the points $M,N,T$ be fixed. Let $P$ be on a fixed (red) line $\ell$ and let $X$ and $Y$ be an ortogonal projections of $P$ on line $MT$ and $TN$. Then the map $X\longmapsto Y$ is a spiral similarity with a center at point $F$ which is an intersection point of circle $PXY$ with $\ell$.

Proof: Since $\angle PFT = \angle PYT =90^{\circ}$ we see that $F$ is a fixed point. Now since $$\angle YXF = \angle YTF =\angle NTF = {\rm fixed \;\; angle}$$ and $$\angle YFX = \angle YTX =\angle NTM = {\rm fixed \;\; angle}$$ the triangle $XYF$ has the same angles for all $P$ so the map $X\longmapsto Y$ is a spiral similarity with rotational angle $\angle MTN$ and center at $F$.

Lemma 2: Say $Z$ is a midpoint of $XY$ from the previous lemma. Then the map $X\longmapsto Z $ is also a spiral similarity (different one) with the same center.

Idea how to finish: Clearly we have a spiral similarity $S_1$ which takes $X\mapsto Y$ and a spiral similarity $S_2$ which takes $Y\mapsto Z$.

So we have $$X\stackrel{S_1}{\longmapsto}Y \stackrel{S_2}{\longmapsto}Z$$ Now take another $X',Y',Z'$:

$$X'\stackrel{S_1}{\longmapsto}Y' \stackrel{S_2}{\longmapsto}Z'$$

I would like to show that there is a spiral similarity which takes $X$ to $X'$, $Y$ to $Y'$ and $Z$ to $Z'$ and this would prove the statement.

Clearly there is a spiral similarity that takes $X$ to $X'$ and $Y$ to $Y'$ induced by $S_1$, but I don't know how to show that it takes also $Z$ to $Z'$.

Actualy, this last claim is not quite true. I know that if $S_1$ and $S_2$ have the same center then we are done, and this is the thing I'm trying to figure out.